Question

\(\dfrac{1}{ab}\ge\dfrac{4}{\left(a+b\right)^2}\)

Answer 1

theo bđt Cô-si A+B ≥ \(2\sqrt{AB}\)

=> \(\dfrac{1}{A}+\dfrac{1}{B}\) ≥ \(2\sqrt{\dfrac{1}{AB}}\)

<=> 2 (\(\dfrac{A+B}{AB}\)) ≥ \(\sqrt{\dfrac{1}{AB}}\)

bình phương cả hai vế

[2(\(\dfrac{A+B}{AB}\))]² ≥ \(\dfrac{1}{AB}\) ≥ \(\sqrt{\dfrac{1}{AB}}\) ∀ A,B > 0

=> \(\dfrac{4\left(A+B\right)^2}{\left(AB\right)^2}\)≥ \(\dfrac{1}{AB}\)

Answer 2

Thiếu đk đó là a,b>0

\(\dfrac{1}{ab}\ge\dfrac{4}{\left(a+b\right)^2}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

\(\Rightarrowđpcm\)