Question

cho a>0,b>0 và ab=1.chứng minh:

\(\dfrac{a}{a^4+b^2}+\dfrac{b}{a^2+b^4}\)\(\le\)1

Answer 1

\(\dfrac{a}{a^4+b^2}+\dfrac{b}{a^2+b^4}\le\dfrac{a}{2\sqrt{a^4b^2}}+\dfrac{b}{2\sqrt{a^2b^4}}=\dfrac{a}{2a^2b}+\dfrac{b}{2ab^2}=\dfrac{1}{ab}=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=1\)