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maxima-minima

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Macedoi
07/08/2017 at 09:26
Answers
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

  • ...
    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

  • ...
    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:


...
Kayasari Ryuunosuke Coordinator
16/07/2017 at 09:44
Answers
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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