Question

chứng minh rằng

\(\dfrac{a^2+b^2}{2}\) ≥ (\(\left(\dfrac{a+b}{2}\right)^2\)

Answer 1

Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)

\(\Leftrightarrow\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow4\cdot\left(a^2+b^2\right)\ge2\cdot\left(a+b\right)^2\)

\(\Leftrightarrow4a^2+4b^2\ge2\cdot\left(a^2+2ab+b^2\right)\)

\(\Leftrightarrow4a^2+4b^2\ge2a^2+4ab+2b^2\)

\(\Leftrightarrow4a^2+4b^2-2a^2-4ab-2b^2\ge0\)

\(\Leftrightarrow2a^2-4ab+2b^2\ge0\)

\(\Leftrightarrow2\left(a^2-2ab+b^2\right)\ge0\)

\(\Leftrightarrow2\left(a-b\right)^2\ge0\)(luôn đúng)