Question

\(\sqrt{a^2+b^2}\ge\dfrac{a+b}{\sqrt{2}}\)

Answer 1

Ta có:
\(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\)\(\Leftrightarrow-a^2-b^2+2ab\le0\Leftrightarrow-a^2-b^2+2ab+2a^2+2b^2\le2a^2+2b^2\)
\(\Leftrightarrow a^2+b^2+2ab\le2a^2+2b^2\)
\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2}\le a^2+b^2\)
\(\Leftrightarrow\dfrac{a+b}{\sqrt{2}}\le\sqrt{a^2+b^2}\)