Question

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Answer 1

\(\left(x-y\right)^2\ge0\Rightarrow x^2+y^2\ge2xy\)

\(\Rightarrow2\left(x^2+y^2\right)\ge x^2+2xy+y^2\)

\(\Rightarrow P\ge\frac{1}{2}\left(x+y\right)^2=\frac{1}{2}\cdot\sqrt{10^2}=5\)

Dấu = khi \(x=y=\frac{\sqrt{10}}{2}\)

Vậy P_min=5 khi \(x=y=\frac{\sqrt{10}}{2}\)