Xét hiệu $\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}$
$=\dfrac{a+b}{ab}-\dfrac{4}{a+b}$
$=\dfrac{(a+b)^2-4ab}{ab(a+b)}$
$=\dfrac{(a-b)^2}{ab(a+b)}$
Mà $(a-b)^2≥0∀a;b$
$ab(a+b)>0∀a;b>0$
$⇒\dfrac{(a-b)^2}{ab(a+b)}≥0$
$⇒\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}≥0$
Hay $\dfrac{1}{a}+\dfrac{1}{b}≥\dfrac{4}{a+b}$
Dấu `=` xảy ra $⇔a-b=0⇔a=b$