Question

Answer 1

Ta có \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)

Dấu = xảy ra <=> x = y

Answer 2

vì \(x+y\ge2\sqrt{xy}\)  (1)  

theo cô si

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\) (2)

lấy (1) nhân (2) theo vế

\(\left(x+y\right)\left(\dfrac{1}{x}\times\dfrac{1}{y}\right)\ge4\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

dấu ''='' xảy ra khi \(x=y\)