Question

bài 2 : cho x,y,z >0 thỏa mãn

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}>\dfrac{1}{\sqrt{xy}+}\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)

Answer 1

\(VT=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{2}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{2}\left(\dfrac{1}{z}+\dfrac{1}{x}\right)\ge\dfrac{1}{2}.\dfrac{2}{\sqrt{xy}}+\dfrac{1}{2}.\dfrac{1}{\sqrt{yz}}+\dfrac{1}{2}.\dfrac{1}{\sqrt{zx}}=\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)

Dấu "=" xảy ra <=> x = y = z > 0