Question

Cho x,y,z>0 thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\). Tìm Max \(P=\frac{1}{\sqrt{2x^2+y^2+3}}+\frac{1}{\sqrt{2y^2+z^2+3}}+\frac{1}{\sqrt{2z^2+x^2+3}}\)

Answer 1

\(\frac{P}{\sqrt{6}}=\sum\frac{1}{\sqrt{6}}.\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{1}{2}\sum\left(\frac{1}{6}+\frac{1}{2x^2+y^2+3}\right)\)

\(\frac{P}{\sqrt{6}}\le\frac{1}{4}+\frac{1}{2}\sum\frac{1}{2\left(x^2+1\right)+\left(y^2+1\right)}\le\frac{1}{4}+\frac{1}{2}\sum\frac{1}{4x+2y}\)

\(\frac{P}{\sqrt{6}}\le\frac{1}{4}+\frac{1}{4}\sum\frac{1}{x+x+y}\le\frac{1}{4}+\frac{1}{36}\left(\frac{2}{x}+\frac{1}{y}+\frac{2}{y}+\frac{1}{z}+\frac{2}{z}+\frac{1}{x}\right)\)

\(\frac{P}{\sqrt{6}}\le\frac{1}{4}+\frac{1}{12}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\)

\(\Rightarrow P\le\frac{\sqrt{6}}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\)