Olympiad Problem 1

Algebra Level 5

Let x,y,z be positive reals for which:

$\sum_{cyc} (xy)^{2}=6xyz$

Find the minimum value of:

$\sum_{cyc} \sqrt{\frac{x}{x+yz}}$

If the minimum value can be expressed in the form a $\sqrt{b}$ + c, then find a + b + c

2 solutions

Ronak Agarwal
Aug 21, 2014

Frist note that :

$\frac { x+yz }{ x } +\frac { y+zx }{ y } +\frac { z+xy }{ z } =\frac { xyz+{ (yz) }^{ 2 }+xyz+{ (zx) }^{ 2 }+xyz+{ (xy })^{ 2 } }{ xyz }$

$=\frac { 3xyz+\sum { { (xy) }^{ 2 } } }{ xyz }$

$\frac { 3xyz+6xyz }{ xyz } =9$

Applying generalised power mean equality we get :

$({ \frac { \sum { \sqrt { \frac { x }{ x+yz } } } }{ 3 } ) }^{ 2 }\ge \frac { 3 }{ \sum { \frac { x+yz }{ x } } }$

From which we directly get :

$\sum { \sqrt { \frac { x }{ x+yz } } } \ge \sqrt { 3 }$

nice solution!

Kartik Sharma - 6 years, 9 months ago

Kinjal Saxena
Oct 3, 2014

The expression can be simplified to

1/(1+yz/X)^.5 +1/(1+zx/y)^.5 +1/(1+xy /z)^.5>=([1+1+1]^2)/A

(note that :A =1/[(1+yz/X)^.5 +(1+zx/y)^.5 +(1+xy/z)^.5])

Also, by cauchy schwartz. Inequality

(1+yz/X)+(1+zx/y)+(1+Xy/z) >=(A ^2)/3

This implies that

1/A>=1/[3.(3+€(xy)^2/xyz)]^.5 (note:€ stands for summation) Or 1/A>=1/(3(3+6xyz/xyz)]^.5

This implies that

(1+1+1)^2/A>=3^.5

Hence the answer is 4.