Question

Cmr \(\sqrt{a^2+b^2}\ge\frac{a+b}{\sqrt{2}}\text{với mọi}a;b\ge0\)

Answer 1

Ta có : \(\sqrt{a^2+b^2}\ge\frac{a+b}{\sqrt{2}}\)

\(\Leftrightarrow a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)( bình phương 2 vế )

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra khi :  \(a=b\)

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