2 replies to this topic

[doctor lee]

cho x,y,z>0 khác nhau đôi một tm (x+z)(y+z)=1

cm $\frac{1}{(x-y)^{2}}+\frac{1}{(x+z)^{2}}+\frac{1}{(y+z)^{2}}\geq 4$

[Tea Coffee]

$a=x+z,b=y+z(a,b> 0)$ $=>ab=1$

CM: $\frac{1}{(a-b)^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\geq 4$

$<=>\frac{1}{(a-b)^{2}}+\frac{(a-b)^{2}}{a^{2}b^{2}}\geq 2<=> \frac{1}{(a-b)^{2}}+(a-b)^{2}\geq 2$ right.

Edited by Tea Coffee, 21-05-2018 - 18:01.

[doctor lee]

..........

......... là gì  vậy ???