Chứng minh S=căn(8/a^2+9b^2/2+c^2a^2/4)+căn...>=6 căn 6

Áp dụng bất đẳng thức Bunyakovsky

\(\Rightarrow\sqrt{\left(\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}\right)\left[\left(\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2+2^2\right]}\ge\left(\sqrt{\dfrac{4}{a}+9b+ca}\right)^2\)

\(\Leftrightarrow2\sqrt{6}\sqrt{\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}}\ge\dfrac{4}{a}+9b+ac\)

Tương tự ta có \(\left\{{}\begin{matrix}2\sqrt{6}\sqrt{\left(\dfrac{8}{b^2}+\dfrac{9c^2}{2}+\dfrac{a^2b^2}{4}\right)}\ge\dfrac{4}{b}+9c+ab\\2\sqrt{6}\sqrt{\left(\dfrac{8}{c^2}+\dfrac{9a^2}{2}+\dfrac{b^2c^2}{4}\right)}\ge\dfrac{4}{c}+9a+bc\end{matrix}\right.\)

\(\Rightarrow2\sqrt{6}S\ge\dfrac{4}{a}+9a+\dfrac{4}{b}+9b+\dfrac{4}{c}+9c+ab+bc+ac\)

\(\Leftrightarrow2\sqrt{6}S\ge\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{a}+a\ge2\sqrt{4}=4\\\dfrac{4}{b}+b\ge2\sqrt{4}=4\\\dfrac{4}{c}+c\ge2\sqrt{4}=4\end{matrix}\right.\)

\(\Rightarrow\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\ge12+8a+8b+8c+ab+bc+ac\)

\(\Rightarrow2\sqrt{6}S\ge12+8a+8b+8c+ab+bc+ac\)

\(\Leftrightarrow2\sqrt{6}S\ge12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow2a+bc\ge2\sqrt{2abc}\)

Tượng tự ta có \(2b+ac\ge2\sqrt{2abc}\) ; \(2c+ab\ge2\sqrt{2abc}\)

\(\Rightarrow12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)

\(\Rightarrow2\sqrt{6}S\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)

Theo đề bài ta có \(a+b+c+\sqrt{2abc}\ge10\)

\(\Rightarrow6\left(a+b+c+\sqrt{2abc}\right)+12\ge72\)

\(\Rightarrow S\ge\dfrac{72}{2\sqrt{6}}=6\sqrt{6}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=2\)