Question

Cho x,y,z là các số dương. Chứng minh rằng:

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

Answer 1

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

\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\) (2)

\(\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xz}}\) (3)

Cộng (1);(2);(3) vế theo vế ta được:

\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)

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