Đáp án:
$ĐKXĐ: x> 0; y>0$
vì $x>0; y>0 => xy >0$
$\dfrac{1}{x} + \dfrac{1}{y} \geq \dfrac{4}{x+y}$
$\dfrac{x+y}{xy} \geq \dfrac{4}{x+y} $
$\frac{x+y}{xy} (x+y)(xy) \geq \frac{4}{x+y}.(xy).(x+y)$
$(x+y)^{2}\geq 4xy$
$=> x^2 +2xy +y^2 \geq 4xy$
$=> x^2-2xy+y^2 \geq 0$
$=>(x-y)^2 \geq 0 $ $\text{( luôn đúng)}$
$=> (Đpcm)$