Question

Chững minh rằng bất đẳng thức : \(\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}.\)

Answer 1

Ta có thể xét hiệu : \(\dfrac{a^2+b^2}{2}-\left(\dfrac{a+b}{2}\right)^2=\dfrac{2\left(a^2+b^2\right)}{4}-\dfrac{\left(a+b\right)^2}{4}\)

\(=\dfrac{2\left(a^2+b^2\right)-\left(a^2+2ab+b^2\right)}{4}\)

\(=\dfrac{1}{4}\left(a^2-2ab+b^2\right)=\dfrac{1}{4}\left(a-b\right)^2\)

Ta thấy : \(\left(a-b\right)^2\ge0\) nên \(\dfrac{1}{4}\left(a-b\right)^2\ge0\)

Hay là : \(\dfrac{a^2+b^2}{2}-\left(\dfrac{a+b}{2}\right)^2\ge0\)

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

=> ĐPCM.