Question

Chứng minh:

\(a^2+\dfrac{b^2}{a^2}+\dfrac{1}{b^2}\ge a+\dfrac{b}{a}+\dfrac{1}{b}a,b\ge0\)

Answer 1

\(a^2+1\ge2a\) ; \(\dfrac{b^2}{a^2}+1\ge\dfrac{2b}{a}\) ; \(\dfrac{1}{b^2}+1\ge\dfrac{2}{b}\)

\(\Rightarrow a^2+\dfrac{b^2}{a^2}+\dfrac{1}{b^2}+3\ge a+\dfrac{b}{a}+\dfrac{1}{b}+a+\dfrac{b}{a}+\dfrac{1}{b}\ge a+\dfrac{b}{a}+\dfrac{1}{b}+3\sqrt[3]{\dfrac{ab}{ab}}\)

\(\Rightarrow a^2+\dfrac{b^2}{a^2}+\dfrac{1}{b^2}+3\ge a+\dfrac{b}{a}+\dfrac{1}{b}+3\)

\(\Rightarrow\) đpcm

Dấu "=" xảy ra khi \(a=b=1\)