Thiếu đk đó là a,b>0
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
Áp dụng bđt AM-GM cho 2 số không âm a,b ta được
\(a+b\ge2\sqrt{ab}\)
TT\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{ab}}\)
Nhân vế theo vế ta được:\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2\sqrt{ab}\cdot2\cdot\sqrt{\dfrac{1}{ab}}=4\left(đpcm\right)\)
xét hiệu
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\ge0\)
⇔ \(\dfrac{b\left(a+b\right)}{ab\left(a+b\right)}+\dfrac{a\left(a+b\right)}{ab\left(a+b\right)}-\dfrac{4ab}{ab\left(a+b\right)}\ge0\)
⇔ \(\dfrac{b\left(a+b\right)+a\left(a+b\right)+4ab}{ab\left(a+b\right)}\ge0\)
⇔\(\dfrac{a^2-2ab+b^2}{ab\left(a+b\right)}\ge0\)
⇔ \(\dfrac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\) (đúng vì a,b>0)
=> đpcm
Ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\) \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( đúng)
