Question

Cho a,b>0 thoả mãn: a+b=1. CMR: \(\dfrac{2}{ab}+\dfrac{3}{a^2+b^2}\ge14\)

Answer 1

Ta có BĐT : \(\left(a+b\right)^2\ge4ab\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}=\dfrac{1}{4}\)

Lại có : \(\dfrac{3}{a^2+b^2}+\dfrac{2}{ab}=\dfrac{9}{3a^2+3b^2}+\dfrac{9}{6ab}+\dfrac{3}{6ab}\)

Theo BĐT Cauchy schwarz ta có :

\(\dfrac{9}{3a^2+3b^2}+\dfrac{9}{6ab}+\dfrac{3}{6ab}\ge\dfrac{\left(3+3\right)^2}{3\left(a+b\right)^2}+\dfrac{3}{6ab}\ge\dfrac{36}{3}+\dfrac{3}{\dfrac{3}{2}}=12+2=14\)

Vậy BĐT đã được chứng minh . Dấu \("="\) xảy ra khi \(a=b=\dfrac{1}{2}\)