Question

[kwspjh]

 \(x+y=4\Rightarrow\frac{x+y}{2}=2\Rightarrow\sqrt{\frac{x+y}{2}}=\sqrt{2}\)

\(P.\sqrt{\frac{x+y}{2}}=\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}\)

\(\Leftrightarrow\sqrt{2}P=\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}\)

\(\Leftrightarrow\sqrt{2}P\ge x+\frac{1}{x}+y+\frac{1}{y}\)

\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge4-3x\)

\(y+\frac{1}{y}=\left(\frac{1}{y}+4y\right)-3y\ge4-3y\)

\(\Rightarrow\sqrt{2}P\ge8-3\left(x+y\right)=8-3.4=-4\)

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