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