Question

[gấukoala]

Cho a,b,c>0. CMR. \(\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}>2\left(\sqrt{\frac{a}{c+b}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}\right)\)

Answer 1

[Victorique de Blois]

có : \(\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\Leftrightarrow x-2\sqrt{xy}+y\ge0\)

\(\Leftrightarrow2x+2y\ge x+2\sqrt{xy}+y\)

\(\Leftrightarrow2\left(x+y\right)\ge\left(\sqrt{x}+\sqrt{y}\right)^2\)

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

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

áp dụng vào ta có :  

\(\sqrt{\frac{a+b}{c}}=\sqrt{\frac{a}{c}+\frac{b}{c}}\ge\frac{1}{\sqrt{2}}\left(\sqrt{\frac{a}{c}}+\sqrt{\frac{b}{c}}\right)\)

\(\sqrt{\frac{b+c}{a}}=\sqrt{\frac{b}{a}+\frac{c}{a}}\ge\frac{1}{\sqrt{2}}\left(\sqrt{\frac{b}{a}}+\sqrt{\frac{c}{a}}\right)\)

\(\sqrt{\frac{c+a}{b}}=\sqrt{\frac{c}{b}+\frac{a}{b}}\ge\frac{1}{\sqrt{2}}\left(\sqrt{\frac{c}{b}}+\sqrt{\frac{a}{b}}\right)\)

\(\Rightarrow VT\ge\frac{1}{\sqrt{2}}\left[\sqrt{a}\left(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)+\sqrt{b}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)+\sqrt{c}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\right]\)

áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có :

\(VT\ge\frac{1}{\sqrt{2}}\left(\sqrt{a}\cdot\frac{4}{\sqrt{b}+\sqrt{c}}+\sqrt{b}\cdot\frac{4}{\sqrt{a}+\sqrt{c}}+\sqrt{c}\cdot\frac{4}{\sqrt{a}+\sqrt{b}}\right)\)

có \(\hept{\begin{cases}\sqrt{b}+\sqrt{c}\le\sqrt{2\left(b+c\right)}\\\sqrt{a}+\sqrt{c}\le\sqrt{2\left(a+c\right)}\\\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\end{cases}}\)  nên \(\hept{\begin{cases}\frac{4}{\sqrt{b}+\sqrt{c}}\ge\frac{4}{\sqrt{2\left(b+c\right)}}\\\frac{4}{\sqrt{a}+\sqrt{c}}\ge\frac{4}{\sqrt{2\left(a+c\right)}}\\\frac{4}{\sqrt{a}+\sqrt{b}}\ge\frac{4}{\sqrt{2\left(a+b\right)}}\end{cases}}\)

\(\Rightarrow vt\ge2\cdot\left(\sqrt{\frac{a}{c+b}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}\right)\)

dấu = xảy ra khi a=b=c

Answer 2

[gấukoala]

sao dòng 9 cs vậy bạn, sao VT lại >=1/ căn 2