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inequality

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Lê Anh Duy
03/05/2019 at 12:58
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35*) Given \(\Delta ABC\). Prove that: \(S_{ABC}\le\dfrac{\sqrt{3}}{12}\left(AB^2+BC^2+CA^2\right)\)

inequalityGeometryarea


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Lê Anh Duy
27/02/2019 at 09:55
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30) Let a, b, c are the length of a triangle. Prove that:

\(\dfrac{ab}{a+b-c}+\dfrac{bc}{b+c-a}+\dfrac{ca}{c+a-b}\ge a+b+c\)

inequality

  • ...
    FacuFeri 28/02/2019 at 15:58

    Put \(A=\dfrac{ab}{a+b-c}+\dfrac{bc}{b+c-a}+\dfrac{ac}{c+a-b}\)

    Because a ; b ; c are the length of a triangle so \(a+b-c;b+c-a;c+a-b>0\)

    Put \(a+b-c=x;b+c-a=y;c+a-b=z\)

    \(\Rightarrow\dfrac{x+y}{2}=b;\dfrac{y+z}{2}=c;\dfrac{x+z}{2}=a;a+b+c=x+y+z\) 

    We have : \(\dfrac{\left(x+y\right)\left(x+z\right)}{2.2x}+\dfrac{\left(x+y\right)\left(y+z\right)}{2.2y}+\dfrac{\left(x+z\right)\left(y+z\right)}{2.2z}\)

    \(=\dfrac{x\left(x+y+z\right)+yz}{4x}+\dfrac{y\left(x+y+z\right)+xz}{4y}+\dfrac{z\left(x+y+z\right)+xy}{4z}\)

    \(=\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{yz}{4x}+\dfrac{xz}{4y}+\dfrac{xy}{4z}\)

    \(=\dfrac{3\left(x+y+z\right)}{4}+\dfrac{y^2z^2}{4xyz}+\dfrac{x^2z^2}{4xyz}+\dfrac{x^2y^2}{4xyz}\)

    Applying the inequality \(a^2+b^2+c^2\ge ab+bc+ac\) , we have :

    \(A\ge\dfrac{3\left(x+y+z\right)}{4}+\dfrac{xyz\left(x+y+z\right)}{4xyz}=x+y+z=a+b+c\)

    Equal sign occurs \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)

     
    Lê Anh Duy selected this answer.
  • ...
    Nguyễn Thị Linh 06/03/2019 at 23:34

    FacuFeri copy internet


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Nguyễn Nhật Minh
02/04/2017 at 22:06
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Prove that \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\left(n\in N;n\ge2\right)\)

Help me please!

Fractioninequality

  • ...
    Nguyễn Huy Tú 03/04/2017 at 13:03

    \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\)

    \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\)

    \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\)

    \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\)

    \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\)

    \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\)

    \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

    So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

    Nguyễn Nhật Minh selected this answer.
  • ...
    AI kết bạn với mình là may mắn cả đời 04/04/2017 at 21:05

    \( \dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

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    Indrace AD-MIN 07/04/2017 at 21:35

    123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14


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Nguyễn Nhật Minh
04/04/2017 at 06:29
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3
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Prove that: 

 \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\)(n \(\in\)N*)

Fractioninequality

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    ¤« 19/02/2018 at 21:29

    34+536+7144+...+2n+1n2(n+1)2

    =1−122+122−132+132−142+...+1(n−1)2+1n2

    =1−1n2<1

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    Lê Quốc Trần Anh Coordinator 12/02/2018 at 09:27

    \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

    \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}\)

    \(=1-\dfrac{1}{n^2}< 1\)

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    FC Alan Walker 21/02/2018 at 17:38

    Ta có: \(\dfrac{2n+1}{n^2\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

    Do đó \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

           \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

           \(=1-\dfrac{1}{\left(n+1\right)^2}< 1\)

    Vậy \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\forall n\in\)N*

                   


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Macedoi
07/08/2017 at 09:26
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10
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Given \(a,b,c\) are non-negative numbers such that \(ab+bc+ca=1\). Find the minimize value of \(P=\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\)

-Source: Câu hỏi của michelle holder - Toán lớp 10 | Học trực tuyến (Ace Legona's solution is wrong)

inequalitymixing-variablemaxima-minima

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    FA KAKALOTS 28/01/2018 at 22:06

    i have a solution but it's ugly

    If a=b=1

     and c=0   then we get a value  2+1√2

    We'll prove that it's a minimal value. Thus, we need to prove that

    √ab+bc+caa2+b2+√ab+bc+cab2+c2+√ab+bc+caa2+c2≥2+1√2

    WLOG c=min{a,b,c}

    . Hence:

    ab+ac+bca2+b2−(a+c)(b+c)(a+c)2+(b+c)2=c(a+b+2c)(2ab+ac+bc)a2+b2)((a+c)2+(b+c)2≥0

    Similarab+ac+bca2+c2−b+ca+c=c(2ab+ac−c2)(a+c)(a2+c2)≥0

    And ab+ac+bcb2+c2−a+cb+c=c(2ab+bc−c2)(b+c)(b2+c2)≥0

    Let a+cb+c=x2;b+ca+c=y2(x,y>0)

    ⇒xy=1

     and we have:

    x+y+1√x2+y2≥2+1√2

    ⇔x+y−2√xy≥1√2−1√x2+y2

    ⇔(√x−√y)2≥(x−y)2√2(x2+y2)(√x2+y2+√2)

    ⇔√2(x2+y2)(√x2+y2+√2)≥(√x+√y)2

    By Cauchy-Schwarz's ine we have: 

    √2(x2+y2)=√(12+12)(x2+y2)≥x+y

    =12(12+12)((√x)2+(√y)2)≥12(√x+√y)2

    Thus, it's enough to prove that √x2+y2+√2≥2

    It's true by AM-GM √x2+y2+√2≥√2xy+√2=2√2>2

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    AL 08/08/2017 at 13:15

    i have a solution but it's ugly

    If \(a=b=1\) and \(c=0\)   then we get a value  \(2+\frac{1}{\sqrt2}\)

    We'll prove that it's a minimal value. Thus, we need to prove that

    \(\sqrt{\dfrac{ab+bc+ca}{a^2+b^2}}+\sqrt{\dfrac{ab+bc+ca}{b^2+c^2}}+\sqrt{\dfrac{ab+bc+ca}{a^2+c^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

    WLOG \(c=\min\{a,b,c\}\). Hence:

    \(\dfrac{ab+ac+bc}{a^2+b^2}-\dfrac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\dfrac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\ge0\)

    Similar\(\dfrac{ab+ac+bc}{a^2+c^2}-\dfrac{b+c}{a+c}=\dfrac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\ge0\)

    And \(\dfrac{ab+ac+bc}{b^2+c^2}-\dfrac{a+c}{b+c}=\dfrac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\ge0\)

    Let \(\dfrac{a+c}{b+c}=x^2;\dfrac{b+c}{a+c}=y^2\left(x,y>0\right)\)\(\Rightarrow xy=1\) and we have:

    \(x+y+\dfrac{1}{\sqrt{x^2+y^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

    \(\Leftrightarrow x+y-2\sqrt{xy}\ge\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{x^2+y^2}}\)

    \(\Leftrightarrow(\sqrt{x}-\sqrt{y})^2\ge\dfrac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})}\)

    \(\Leftrightarrow\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})\ge(\sqrt{x}+\sqrt{y})^2\)

    By Cauchy-Schwarz's ine we have: 

    \(\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\ge x+y\)

    \(=\dfrac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\ge\dfrac{1}{2}(\sqrt{x}+\sqrt{y})^2\)

    Thus, it's enough to prove that \(\sqrt{x^2+y^2}+\sqrt{2}\ge2\)

    It's true by AM-GM \(\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2\)

     
  • ...
    Phan Huy Toàn 20/08/2017 at 20:10

    tôi để lại số quả táo là:


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Aim Egst
25/07/2017 at 19:04
Answers
2
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Given \(m\ge n>0\) and \(a,b,c\) are positive real numbers. Prove this inequality \(\dfrac{a^m}{b^n+c^n}+\dfrac{b^m}{c^n+a^n}+\dfrac{c^m}{a^n+b^n}\ge\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{2}\)

 it's a general formula of Nesbitt's inequality, i have a method but i need more :)), Help me, thanks 

 

inequality

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    FA KAKALOTS 28/01/2018 at 22:06

    My try: WLOG a≥b≥c hence we have: am−n≥bm−n≥cm−n

    ⇒anbn+cn≥bncn+an≥cnan+bn

    By Chebyshev's inequality we have: 

    ∑ambn+cn=∑am−n⋅anbn+cn

    ≥am−n+bm−n+cm−n3⋅∑anbn+cn

    ≥∑am−n+bm−n+cm−n2

    We're done when a=b=c

    P/s: This's discuss not A4 (Auto Ask Auto Answer)

  • ...
    Aim Egst 25/07/2017 at 19:12

    My try: WLOG \(a\ge b\ge c\) hence we have: \(a^{m-n}\ge b^{m-n}\ge c^{m-n}\)

    \(\Rightarrow\dfrac{a^n}{b^n+c^n}\ge\dfrac{b^n}{c^n+a^n}\ge\dfrac{c^n}{a^n+b^n}\)

    By Chebyshev's inequality we have: 

    \(\text{∑}\dfrac{a^m}{b^n+c^n}=\text{∑}a^{m-n}\cdot\dfrac{a^n}{b^n+c^n}\)

    \(\ge\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{3}\cdot\text{∑}\dfrac{a^n}{b^n+c^n}\)

    \(\ge\text{∑}\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{2}\)

    We're done when \(a=b=c\)

    P/s: This's discuss not A4 (Auto Ask Auto Answer)


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Macedoi
24/07/2017 at 18:42
Answers
1
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Given $a,b,c>0$. Prove that $\frac{a^3b}{3a+b}+\frac{b^3c}{3b+c}+\frac{c^3a}{3c+a}\ge \frac{a^2bc}{2a+b+c}+\frac{ab^2c}{a+2b+c}+\frac{abc^2}{a+b+2c}$

*)Source: https://olm.vn/hoi-dap/question/999979.html

I tried AM-GM but unsuccess 

inequality

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    Ely Seon 05/08/2018 at 09:54

    Considering the right expression we have  \(\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}\)

    \(=abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\)

    Apply Cauchy Schwarz inequality we have the formula \(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{1}{a+b}\)

    Apply this formula to the following expression we have

    \(\dfrac{a}{a+b+a+c}\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)\(;\dfrac{b}{a+b+b+c}\le\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)

    \(\dfrac{c}{a+c+c+a}\le\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{c+a}\right)\)

    \(\Rightarrow\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+c+a}\le\dfrac{a+b}{4\left(a+b\right)}+\dfrac{b+c}{4\left(b+c\right)}+\dfrac{c+a}{4\left(c+a\right)}=\dfrac{3}{4}\)

    \(\Rightarrow abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}\)

    We now need to prove that \(\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}\)

    \(\Leftrightarrow\dfrac{a^3b}{3a+b}-\dfrac{abc}{4}+\dfrac{b^3c}{3b+c}-\dfrac{abc}{4}+\dfrac{c^3a}{3c+a}-\dfrac{abc}{4}\ge0\)

    \(\Leftrightarrow\dfrac{a^3bc}{c\left(3a+b\right)}-\dfrac{abc}{4}+\dfrac{ab^3c}{a\left(3b+c\right)}-\dfrac{abc}{4}+\dfrac{c^3ab}{b\left(3c+a\right)}-\dfrac{abc}{4}\ge0\)

    \(\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{b^2}{a\left(3b+c\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{1}{4}\right]\ge0\)

    \(\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{3}{4}\right]\ge0\)

    \(\Leftrightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}\)

    Apply Cauchy Schwarz inequality Engel form we have 

    \(\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}\)

    According to the Cauchy's consequence, we have \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

    \(\Rightarrow\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}\ge\dfrac{3\left(ab+bc+ac\right)}{4\left(ab+bc+ac\right)}=\dfrac{3}{4}\)

    \(\Rightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}\)

    So now we have \(\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}\)

    But \(abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}\)

    \(\Rightarrow\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}\) ( things must be proven.)


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Aim Egst
18/07/2017 at 18:21
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0
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For \(a,x,y,z\) are positive real numbers satisfy \(xyz=1\) and \(a\ge 1\). Prove that \(\dfrac{x^a}{y+z}+\dfrac{y^a}{x+z}+\dfrac{z^a}{x+y}\ge\dfrac{3}{2}\)

-Like Nesbitt's inequality, i have a problem it's a general formula of Nesbitt's inequality,  if necessary please inbox to me :)

 

 

inequality


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Ace Legona
16/07/2017 at 12:57
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4
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Prove that with \(n>3\) and \(x_1;x_2;...;x_n>0\) satisfy \(\Pi^n_{i=1}x_i=1\), so \(\dfrac{1}{1+x_1+x_1x_2}+\dfrac{1}{1+x_2+x_2x_3}+...+\dfrac{1}{1+x_n+x_nx_1}>1\)

- Source: The problem from Russia,2004

P/s: Spammer go away, Contraindication for young buffalo !!

 

 

inequality

  • ...
    FA KAKALOTS 28/01/2018 at 22:06

    Yes but it's not mine, i need new method :)

    Let x1=a2a1;x2=a3a2;...;xn=a1an

    We need prove a1a1+a2+a3+a2a2+a3+a4+...+anan+a1+a2>1

    It's right because n>3

     and ai+ai+1+ai+2<a1+a2+...+an∀i

  • ...
    Aim Egst 18/07/2017 at 11:32

    Yes but it's not mine, i need new method :)

    Let \(x_1=\dfrac{a_2}{a_1};x_2=\dfrac{a_3}{a_2};...;x_n=\dfrac{a_1}{a_n}\)

    We need prove \(\dfrac{a_1}{a_1+a_2+a_3}+\dfrac{a_2}{a_2+a_3+a_4}+...+\dfrac{a_n}{a_n+a_1+a_2}>1\)

    It's right because \(n>3\) and \(a_i+a_{i+1}+a_{i+2}< a_1+a_2+...+a_n\forall i\)

  • ...
    Summer Clouds moderators 18/07/2017 at 08:33

    Ace Legona : you can share a answer of problem.


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Kayasari Ryuunosuke Coordinator
16/07/2017 at 09:44
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3
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Give a,b,c are positive real number and a2 + b2 + c2 = 3

Find MaxP know :

P = a + b + c - abc 

 

inequalitymaxima-minima

  • ...
    Vũ Mạnh Hùng 05/12/2017 at 20:04

    P=0 vì a+b+c=abc

    suy ra abc -abc=0

    answer :0

  • ...
    Vũ Mạnh Hùng 05/12/2017 at 19:59

    P=0 vì a+b+c=abc

    suy ra abc -abc=0

    answer :0

  • ...
    Ace Legona 16/07/2017 at 12:46

    wrong tag : maxima-minima


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Nguyễn Huy Thắng
09/03/2017 at 20:58
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1
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Let \(a,b,c\) are positive real number satisfy \(a+b+c=1\). Prove that \(\dfrac{1}{\sqrt{\left (a^2+ab+b^2\right )\left (b^2+bc+c^2\right )}}+\dfrac{1}{\sqrt{\left (b^2+bc+c^2\right )\left (c^2+ca+a^2\right )}}+\dfrac{1}{\sqrt{\left (c^2+ca+a^2\right )\left (a^2+ab+b^2\right )}}\ge 4+\dfrac{8}{\sqrt{3}}\)

inequality

  • ...
    Nguyen Huu Ai Linh 10/12/2017 at 07:08

    That too long!


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Nguyễn Nhật Minh
04/04/2017 at 18:29
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0
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Prove that:

B = \(\dfrac{1}{2!}+\dfrac{5}{3!}+\dfrac{11}{4!}+...+\dfrac{n^2+n-1}{\left(n+1\right)!}< 2\).

Help me please! This is my homework and I must give it to my teacher tomorrow!

inequalityFraction


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Nguyễn Nhật Minh
02/04/2017 at 11:25
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2
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Prove that E = \(\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{n^3}< \dfrac{1}{12}\).

Fractioninequality

  • ...
    FA KAKALOTS 28/01/2018 at 22:08

    For any natural number n > 1,we have :

    (n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

    ⇒1n3<1(n−1)n(n+1)

    1(n−1)n(n+1)=1n.1(n−1)(n+1)

    =1n.(n+1)−(n−1)(n−1)(n+1).12=12.1n.(1n−1−1n+1)

    =12.(1(n−1)n−1n(n+1))

    Now we have :

    E < 12.3.4+13.4.5+14.5.6+...+1(n−1)n(n+1)

    =12(12.3−13.4)+12(13.4−14.5)+12(14.5−15.6)+...+12(1(n−1)n−1n(n+1))

    =12(12.3−1n(n+1))=112−12n(n+1)<112

    Hence,E<112

  • ...
    Phan Thanh Tinh Coordinator 24/04/2017 at 13:50

    For any natural number n > 1,we have :

    (n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

    \(\Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

    \(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{n}.\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

    \(=\dfrac{1}{n}.\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}.\dfrac{1}{2}=\dfrac{1}{2}.\dfrac{1}{n}.\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

    \(=\dfrac{1}{2}.\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

    Now we have :

    E < \(\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

    \(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+\dfrac{1}{2}\left(\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)+\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

    \(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{12}-\dfrac{1}{2n\left(n+1\right)}< \dfrac{1}{12}\)

    Hence,\(E< \dfrac{1}{12}\)


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