Áp dụng BĐT AM-GM,ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{4\left(x+y\right)}{x+y}\)
\(\Leftrightarrow\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge4\) ( đfcm )
Có: \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge4\)⇔\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)⇔\(\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\)
⇔\(\dfrac{\left(x+y\right)\left(x+y\right)}{xy\left(x+y\right)}\ge\dfrac{4xy}{xy\left(x+y\right)}\)⇔\(\left(x+y\right)^2\ge4xy\)⇔\(x^2+2xy+y^2\ge4xy\)
⇔\(x^2-4xy+2xy+y^2\ge0\)⇔\(x^2-2xy+y^2\ge0\)⇔\(\left(x-y\right)^2\ge0\) luôn đúng
