Question

Cho a,b>0 thỏa mãn: a+b=1. CM: \(\left(a+\dfrac{1}{a}\right).\left(b+\dfrac{1}{b}\right)\ge\dfrac{25}{4}\)

Answer 1

Ta có : \(\left(a+\dfrac{1}{a}\right)\left(b+\dfrac{1}{b}\right)=ab+\dfrac{1}{ab}+\dfrac{a}{b}+\dfrac{b}{a}\)

\(=\left(ab+\dfrac{1}{16ab}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{15}{16ab}\)

Áp dụng BĐT Cô - si có 

\(ab+\dfrac{1}{16ab}\ge2\sqrt{ab\cdot\dfrac{1}{16ab}}=\dfrac{1}{2}\)

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

Có : \(1=a+b\ge2\sqrt{ab}\Rightarrow ab\le\dfrac{1}{4}\Rightarrow16ab\le4\Rightarrow\dfrac{15}{16ab}\ge\dfrac{15}{4}\)

Do đó \(\left(a+\dfrac{1}{a}\right)\left(b+\dfrac{1}{b}\right)\ge2+\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{25}{4}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)