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![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT\ge\frac{4\left(\sum\sqrt{a}\right)^2}{2\sum\sqrt{a}}=2\sum\sqrt{a}=VP\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(\frac{a}{b}+\frac{a}{c}+\frac{c}{b}+\frac{c}{a}+\frac{b}{c}+\frac{b}{a}\ge\sqrt{2}\left(\Sigma\sqrt{\frac{1-a}{a}}\right)\)
or \(\Sigma\frac{b+c}{a}\ge\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\)
Theo AM-GM:\(\frac{b+c}{a}\ge2\sqrt{\frac{2\left(b+c\right)}{a}}-2\)
Tương tự và cộng lại: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-6\)
Mà: \(\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\ge3\sqrt[6]{\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}}\ge6\)
Từ đó: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}=VP\)
Done!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{2}{\sqrt{a}}+\frac{2}{\sqrt{b}}+\frac{2}{\sqrt{c}}=\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)+\left(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)+\left(\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}\right)\)
\(\ge\frac{\left(1+1\right)^2}{\sqrt{a}+\sqrt{b}}+\frac{\left(1+1\right)^2}{\sqrt{b}+\sqrt{c}}+\frac{\left(1+1\right)^2}{\sqrt{c}+\sqrt{a}}\)
\(=\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{4}{\sqrt{b}+\sqrt{c}}+\frac{4}{\sqrt{c}+\sqrt{a}}\)
=> \(2\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)\(\ge4\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)
=> \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)\(\ge2\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)
"=" xảy ra <=> a =b =c.
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét: \(\sqrt{\frac{a}{b+c+d}}=\frac{\sqrt{a}}{\sqrt{b+c+d}}=\frac{a}{\sqrt{a\left(b+c+d\right)}}\)
\(\sqrt{\frac{b}{c+d+a}}=\frac{\sqrt{b}}{\sqrt{c+d+a}}=\frac{b}{\sqrt{b\left(c+d+a\right)}}\)
\(\sqrt{\frac{c}{d+a+b}}=\frac{\sqrt{c}}{\sqrt{d+a+b}}=\frac{c}{\sqrt{c\left(d+a+b\right)}}\)
\(\sqrt{\frac{d}{a+b+c}}=\frac{\sqrt{d}}{\sqrt{a+b+c}}=\frac{d}{\sqrt{d\left(a+b+c\right)}}\)
\(\Rightarrow VT=\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{a\left(b+c+d\right)}\le\frac{a+b+c+d}{2}\\\sqrt{b\left(c+d+a\right)}\le\frac{a+b+c+d}{2}\\\sqrt{c\left(d+a+b\right)}\le\frac{a+b+c+d}{2}\\\sqrt{d\left(a+b+c\right)}\le\frac{a+b+c+d}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{a\left(b+c+d\right)}}\ge\frac{2a}{a+b+c+d}\\\frac{b}{\sqrt{b\left(c+d+a\right)}}\ge\frac{2b}{a+b+c+d}\\\frac{c}{\sqrt{c\left(d+a+b\right)}}\ge\frac{2c}{a+b+c+d}\\\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge\frac{2d}{a+b+c+d}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}\)
\(\Rightarrow VT\ge\frac{2\left(a+b+c+d\right)}{a+b+c+d}\)
\(\Rightarrow VT\ge2\)
\(\Rightarrow\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge2\)
\(\Leftrightarrow\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{d+a+b}}+\sqrt{\frac{d}{a+b+c}}\ge2\) ( đpcm )
Lời giải:
Áp dụng bất đẳng thức AM-GM:
\(\frac{b+c+d}{a}=\frac{b+c+d}{a}.1\leq \left(\frac{\frac{b+c+d}{a}+1}{2}\right)^2=\left(\frac{b+c+d+a}{2a}\right)^2\)
\(\sqrt{\frac{a}{b+c+d}}\geq \frac{2a}{a+b+c+d}\). Tương tự với các phân thức còn lại:
\(\Rightarrow \text{VT}\geq \frac{2(a+b+c+d)}{a+b+c+d}=2\) (đpcm)
Ta sử dụng bất đẳng thức Chebyshev sau đây:
Nếu các số \(a\ge b\ge c,x\ge y\ge z\) thì \(3\left(ax+by+cz\right)\ge\left(a+b+c\right)\left(x+y+z\right).\)
Thực vậy bất đẳng thức cần chứng minh tương đương với \(\left(a-b\right)\left(x-y\right)+\left(b-c\right)\left(y-z\right)+\left(c-a\right)\left(z-x\right)\ge0.\)
Không mất tính tổng quát, giả sử \(a\ge b\ge c\). Khi đó bất đẳng thức cần chứng minh tương đương với
\(\frac{a+b}{\sqrt{c\left(a+b\right)}}+\frac{b+c}{\sqrt{a\left(b+c\right)}}+\frac{c+a}{\sqrt{b\left(c+a\right)}}\ge2\left(\frac{c}{\sqrt{c\left(a+b\right)}}+\frac{a}{\sqrt{a\left(b+c\right)}}+\frac{b}{\sqrt{b\left(c+a\right)}}\right)\)
\(\leftrightarrow\frac{a+b-2c}{\sqrt{c\left(a+b\right)}}+\frac{c+a-2b}{\sqrt{b\left(c+a\right)}}+\frac{b+c-2a}{\sqrt{a\left(b+c\right)}}\ge0\) (***)
Tuy nhiên ta có \(a+b-2c\ge c+a-2b\ge b+c-2a\) và \(\frac{1}{\sqrt{c\left(a+b\right)}}\ge\frac{1}{\sqrt{b\left(c+a\right)}}\ge\frac{1}{\sqrt{a\left(b+c\right)}}\) nên theo bất đẳng thức Chebyshev
\(\frac{a+b-2c}{\sqrt{c\left(a+b\right)}}+\frac{c+a-2b}{\sqrt{b\left(c+a\right)}}+\frac{b+c-2a}{\sqrt{a\left(b+c\right)}}\)
\(\ge\frac{1}{3}\left(a+b-2c+b+c-2a+c+a-2b\right)\left(\frac{1}{\sqrt{c\left(a+b\right)}}+\frac{1}{\sqrt{b\left(c+a\right)}}+\frac{1}{\sqrt{a\left(b+c\right)}}\right)=0.\)
Vậy bất đẳng thức (***) đúng, nên ta có điều phải chứng minh.