Question

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

Answer 1

có: (a-b)²\(\ge\)0

=>\(a^2-2ab+b^2\ge0\)

=>\(a^2+b^2\ge2ab\)=> a²+b²+a²+b²\(\ge\)a²+b²+2ab

=>\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

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

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

Answer 2

Ta có:(a-b)²\(\ge\)0 với mọi a,b

=>a²-2ab+b²\(\ge\)0 với mọi a,b

=>a²+b²\(\ge\)2ab với mọi ab

=>2(a²+b²)\(\ge\)a²+2ab+b²

=>a²+b²\(\ge\)\(\dfrac{\left(a+b\right)^2}{2}\)

=>\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}=\left(\dfrac{a+b}{2}\right)^2\)(đpcm)