\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}< =>\frac{a+b}{ab}\ge\frac{4}{a+b}< =>\left(a+b\right)^2\ge4ab< =>\left(a-b\right)^2\ge0\left(lđ\right).\)
Dấu "=" xảy ra khi a=b
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{b\left(a+b\right)+a\left(a+b\right)-4ab}{ab\left(a+b\right)}\ge0\)
\(\Leftrightarrow\frac{a^2-2ab+b^2}{ab\left(a+b\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\)(luon dung)
Với a, b > 0 ,theo bất đẳng thức cosi:
\(\frac{1}{a}\)+\(\frac{1}{b}\)\(\ge\)2\(\sqrt{\frac{1}{a}\cdot\frac{1}{b}}\)=\(\frac{2}{\sqrt{ab}}\)\(\ge\)\(\frac{2}{\frac{a+b}{2}}\)=\(\frac{4}{a+b}\)(dpcm)
use cauchy-schwarz inequality,have : 1/a + 1/b >= (1^2+1^2)^2/(a+b)=4/(a+b)
