Question

Cho a,b dương CMR 

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

Answer 1

các bạn giải nhanh giúp mk vs 

Answer 2

BĐT<=> 

\(\left(\frac{2ab}{a+b}-\frac{a+b}{2}\right)+\left(\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\right)\ge0\)

<=> \(-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\frac{a^2+b^2}{2}-ab}{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}\ge0\)

<=> \(\frac{\left(a-b\right)^2}{2(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab})}-\frac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)

<=> \(a+b\ge\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\)

<=> \(\frac{a^2+b^2}{2}+ab\ge2\sqrt{\frac{a^2+b^2}{2}.ab}\)luôn đúng

=> ĐPCM

Dấu bằng xảy ra khi a=b