inequality
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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 \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)
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Nguyễn Thị Linh 06/03/2019 at 23:34
FacuFeri copy internet
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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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\(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}\)
\(=1-\dfrac{1}{n^2}< 1\)
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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*
Macedoi
07/08/2017 at 09:26
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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 \(\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2\)
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Phan Huy Toàn 20/08/2017 at 20:10
tôi để lại số quả táo là:
Aim Egst
25/07/2017 at 19:04
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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)
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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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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.)
Aim Egst
18/07/2017 at 18:21
Ace Legona
16/07/2017 at 12:57
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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
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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 \(a_i+a_{i+1}+a_{i+2}< a_1+a_2+...+a_n\forall i\)
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Ace Legona : you can share a answer of problem.
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Ace Legona 16/07/2017 at 12:46
wrong tag : maxima-minima
Nguyễn Huy Thắng
09/03/2017 at 20:58
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Nguyen Huu Ai Linh 10/12/2017 at 07:08
That too long!
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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
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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,\(E< \dfrac{1}{12}\)