Math-lover123 wrote: Let $x,y,z$ be positive reals for which: $\sum (xy)^{2}=6xyz$ Prove that: $\sum \sqrt{\frac{x}{x+yz}}\geq \sqrt{3}$. After homogenization and substitution $x^2y^2=c$, $x^2z^2=b$ and $y^2z^2=a$ we need to prove that: $\sum_{cyc}\sqrt{\frac{a+b+c}{a+b+7c}}\geq\sqrt3$, which is obvious by Holder.

Another solution: We have that $ \sum\frac{xy}{z}=6$ and we aim to prove that $ \sum\sqrt{\frac{1}{1+\frac {xy}{z}}}\geq\sqrt{3} $.Then notice that $f(x)=\sqrt{\frac{1}{1+x}}$ is convex and use Jensen.

If we say ${ \sqrt{\frac{x}{x+yz}}} =a $ then ${1-a^2\over a^2} = {yz\over x}$ ${ \sqrt{\frac{y}{y+zx}} }=b $ then ${1-b^2\over b^2} = {zx\over y}$ ${ \sqrt{\frac{z}{z+xy}}} =c $ then ${1-c^2\over c^2} = {xy\over z}$ So ${1-a^2\over a^2} +{1-b^2\over b^2}+{1-c^2\over c^2} = 6$ . Now we have to prove $a+b+c \geq \sqrt{3}$ if ${1\over a^2} +{1\over b^2}+{1\over c^2} = 9$. Now if we take A.M.-G.M. we have: \[(a+b+c)^2\cdot 9 =(a^2+b^2+c^2+ab+ab+bc+bc+ac+ac)({1\over a^2} +{1\over b^2}+{1\over c^2}) \geq 27\] and we are done.

Lyub4o wrote: Another solution: We have that $ \sum\frac{xy}{z}=6$ and we aim to prove that $ \sum\sqrt{\frac{1}{1+\frac {xy}{z}}}\geq\sqrt{3} $.Then notice that $f(x)=\sqrt{\frac{1}{1+x}}$ is convex and use Jensen. Very nice! Congratulations!

Another solution: Suppose $ \frac{yz}{x}=a, \frac{zx}{y}=b, \frac{xy}{z}=c $. Then $ a+b+c=6 $ and the inequality is $ S= \sum_{cyc} \frac{1}{\sqrt{1+a}} \ge \sqrt3 $. By titu's lemma, $ S= \sum_{cyc} \frac{1}{\sqrt{1+a}} \ge \frac{9}{\sum_{cyc} \sqrt{1+a}} $. But $ 3(1+a+1+b+1+c)=27 \ge (\sum_{cyc} \sqrt{1+a})^2 $ by cauchy schwarz. Thus $ S \ge \frac{9}{\sqrt{27}}=\sqrt3 $ as desired.

Math-lover123 wrote: Let $x,y,z$ be positive reals for which: $\sum (xy)^{2}=6xyz$ Prove that: $\sum \sqrt{\frac{x}{x+yz}}\geq \sqrt{3}$. let $\frac{yz}{x}=a,\frac{zx}{y}=b,\frac{xy}{z}=c$ than $a+b+c=6$ and $\sum \sqrt{\frac{x}{x+yz}}=\sum \sqrt{\frac{1}{1+a}}$ so by holder's we have $(\sum \sqrt{\frac{1}{1+a}})^{2/3}(\sum 1+a)^{1/3}\ge \sum a\Longrightarrow \sum \sqrt{\frac{1}{1+a}}\ge \sqrt 3$ as $a+b+c=6$ so we are done

aditya21 wrote: Math-lover123 wrote: Let $x,y,z$ be positive reals for which: $\sum (xy)^{2}=6xyz$ Prove that: $\sum \sqrt{\frac{x}{x+yz}}\geq \sqrt{3}$. let $\frac{yz}{x}=a,\frac{zx}{y}=b,\frac{xy}{z}=c$ than $a+b+c=6$ and $\sum \sqrt{\frac{x}{x+yz}}=\sum \sqrt{\frac{1}{1+a}}$ so by holder's we have $(\sum \sqrt{\frac{1}{1+a}})^{2/3}(\sum 1+a)^{1/3}\ge \sum a\Longrightarrow \sum \sqrt{\frac{1}{1+a}}\ge \sqrt 3$ as $a+b+c=6$ so we are done Nice solution...

After using some basic inequalities, we should prove that 1/(x+4)+1/(y+4)+1/(z+4)>=1/2, for x=ab/c, y=bc/a, z=ac/b and from the condition x+y+z=6. Just AM-HM. This problem is nice.