Prove: $$\frac{2x^2+xy}{(y+\sqrt{zx}+z)^2}+\frac{2y^2+yz}{(z+\sqrt{xy}+x)^2}+\frac{2z^2+zx}{(x+\sqrt{yz}+y)^2}\ge1$$ ($x,y,z>0$)
Asked
Active
Viewed 2,568 times
5
-
Schwarz, Chebyshev :( – Tĩnh Thu Feb 19 '14 at 13:42
-
Just like that, without any manipulation to the main statement? – Hawk Feb 19 '14 at 13:44
-
I don'know how to ... – Tĩnh Thu Feb 19 '14 at 13:56
-
Are $x,y,z$ all non-negative, or just real numbers? – Yiyuan Lee Feb 19 '14 at 13:58
-
@YiyuanLee That doesn't matter actually, if one of them is negative, then the denominator part in one of the left hand inequality will turn imaginery. So, they are positive. – Hawk Feb 19 '14 at 14:02
-
@YiyuanLee So,either all of them are positive or all of them are negative, – Hawk Feb 19 '14 at 14:03
-
@TĩnhThu Are you sure that the question is not missing any context? – Hawk Feb 19 '14 at 14:12
1 Answers
6
Using Cauchy-Schwarz,for each expression :
$\left ( y+\sqrt{xz}+z \right )^{2}\le \left ( y+x+x \right )\left ( y+z+\frac{z^{2}}{x} \right )=\dfrac{\left ( 2x^{2}+xy \right )\left ( xy+xz+z^{2} \right )}{x^{2}}$
We have,
$\dfrac{2x^2+xy}{(y+\sqrt{zx}+z)^2}+\dfrac{2y^2+yz}{(z+\sqrt{xy}+x)^2}+\dfrac{2z^2+zx}{(x+\sqrt{yz}+y)^2}\ge \sum_{cyclic} \dfrac{x^{2}}{xy+xz+z^{2}}$
and again using Cauchy-Schwarz :
$\sum_{cyclic} \dfrac{x^2}{xy+xz+z^2}\ge \dfrac{(x+y+z)^2}{2(xy+yz+zx)+x^2+y^2+z^2}=\dfrac{(x+y+z)^2}{(x+y+z)^2}=1$. Q.E.D.
r9m
- 17,938