too simple... let's make the following substitution: $ x=\sin^2{\alpha}$ and so on... it follows that $ 2\sin{\alpha}\sin{\beta}\sin{\gamma}=\sum \cos{\alpha}\sin{\alpha}$ But it means that $ \alpha+\beta+\gamma=\pi$,then obviously $ (\sin{\alpha}\sin{\beta}\sin{\gamma})^2\leq \frac{27}{64}$.

When the equality holds?

when $ x=y=z=\frac{3}{4}$.

Can't we prove the inequality without triangular substituition?

Why from $ 2\sin{\alpha}\sin{\beta}\sin{\gamma}=\sum \cos{\alpha}\sin{\alpha}$ we know that this angles are angles in some triangle(acute angled probably).?

matex (L)(L)(L) wrote: Why from $ 2\sin{\alpha}\sin{\beta}\sin{\gamma} = \sum \cos{\alpha}\sin{\alpha}$ we know that this angles are angles in some triangle(acute angled probably).? actually,this is a well-known identity,I'm pretty sure that the proof is available in any rational geometry book,for example,Prasolov. anyway I would do that by making the following transformations(s): $ 2\sum\cos{\alpha}\sin{\alpha}=\sum \sin{2\alpha}=2\sin\gamma(\cos{(\alpha-\beta)}+\cos\gamma)=4\prod\sin{\gamma}$

kunny wrote: Can't we prove the inequality without triangular substituition? Indeed, we can, and more importantly, the suggested solution from the organizing committee was with NO trigonometric substitution. So, is there any non-trigo answers?

KenHungKK wrote: Given $ x,y,z\in (0,1)$ satisfying that $ \sqrt {\frac {1 - x}{yz}} + \sqrt {\frac {1 - y}{xz}} + \sqrt {\frac {1 - z}{xy}} = 2$. Find the maximum value of $ xyz$. We have that \[ 2 \sqrt {xyz} = \frac {1}{\sqrt {3}} \sum{\sqrt {x(3 - 3x)}} \leq \frac {1}{\sqrt {3}} \sum \frac {x + 3(1 - x)}{2} = \] \[ = \frac {3\sqrt {3}}{2} - \frac {1}{\sqrt {3}} \sum{x}. \] So $ 2 \sqrt {xyz} \leq \frac {3\sqrt {3}}{2} - \frac {1}{\sqrt {3}} \sum{x} \leq \frac {3\sqrt {3}}{2} - \sqrt {3} \cdot \sqrt [3]{xyz}.$ If we denote $ p = \sqrt [6]{xyz}$ we get that $ 2p^3 \leq \frac {3\sqrt {3}}{2} - \sqrt {3}p^2$. This is equivalent to \[ 4p^3 + 2\sqrt {3}p^2 - 3\sqrt {3} \leq 0 \Rightarrow (2p - \sqrt {3})(2p^2 + 2\sqrt {3}p + 3) \leq 0, \] then $ p \leq \frac {\sqrt {3}}{2}$. So $ xyz \leq \frac {27}{64}$. The equality holds for $ x = y = z = \frac {3}{4}$.

Ya the one turcas_c posted is the official solution and..... I think it's still natural. I in fact thought of using AM-GM in this way but then I failed to use it for a second time as the solution does. But then I would say the trigo solution is more beautiful... which I still remember we Hong Kong team solved it in this method on the coach when we're heading to visit the Waterfall LOL P.S. turcas_c i think you're one of the Romanian Team members participating in this competition right?

brianchung11 wrote: P.S. turcas_c i think you're one of the Romanian Team members participating in this competition right? Yap, I am the guy that you (I think) were trying to explain how to solve the rubik cube ), and also the one that you gave your e-mail adress. I hope that we'll keep in touch.

i also solved it using the trig substitution but i cant prove the fact that if the identity is true for some angles $ A,B,C$ then $ A+B+C=180$ can anyone provide the proof clearly?

Given $ x,y,z\in (0,1)$ satisfying that$ \sqrt{\frac{1 - x}{yz}} + \sqrt{\frac{1 - y}{xz}} + \sqrt{\frac{1 - z}{xy}} = 2$.Find the maximum value of $ x^4y^5z^6$. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=518033

The conclusion is simple and the known condition is complex. So we can consider to prove it by contradiction. This is a way without triangular substitution. Firstly, guess the equality holds for $x=y=z=\frac{3}{4}$. So the maximum value of $xyz$ is $\frac{27}{64}$ . Suppose that $xyz>\frac{27}{64}$. Then we only need to prove $\sum\limits_{cyc}\sqrt{\frac{1-x}{yz}}<2.$ $\therefore \sum\limits_{cyc}\sqrt{\frac{1-x}{yz}}=\sum\limits_{cyc}\frac{\sqrt{x(1-x)}}{\sqrt{xyz}}<\frac{\sum\limits_{cyc}\sqrt{x(1-x)}}{\sqrt{\frac{27}{64}}}=\frac{8}{3\sqrt{3}}\sum\limits_{cyc}\sqrt{x(1-x)}$ Thus we only need to prove that $\sum\limits_{cyc}\sqrt{x(1-x)}\le\frac{3\sqrt{3}}{4}$ According to Cuachy Inequality: $ \sum\limits_{cyc}\sqrt{x(1-x)}\le \sqrt{(x+y+z)((1-x)+(1-y)+(1-z))}=\sqrt{s(3-s)}=\sqrt{-{(s-\frac{3}{2})}^2+\frac{9}{4}}.$ Where $s=x+y+z.$ According to AM-GM Inequality: $s\ge 3\sqrt[3]{xyz} >\frac{9}{4}>\frac{3}{2}.$ $ \therefore {(s-\frac{3}{2})}^2>{(\frac{9}{4}-\frac{3}{2})}^2=\frac{9}{16}\Longrightarrow \sum\limits_{cyc}\sqrt{x(1-x)}\le \sqrt{-{(s-\frac{3}{2})}^2+\frac{9}{4}}<\sqrt{-\frac{9}{16}+\frac{9}{4}}=\frac{3\sqrt{3}}{4}.$ $\therefore \sum\limits_{cyc}\sqrt{\frac{1-x}{yz}}<2\Longrightarrow$ contradictory to condition !$\Longrightarrow xyz\le\frac{27}{64}.$ $Q.E.D.$