has anyone have an idea what to do here? Leo Telosa, my team mate from the Philippines and i had the same idea/ draft solution. But this idea/ draft solution of ours was something we invented and had a few assumptions i'm sure we're not allowed to make. I just want to have a starting point for this thing. Btw, the only training in Trig i had was the day before the contest, make that the night before the contest... whatta crammer. Anyway, what we did was since the equation we had to prove had 3 addends, we decided to prove that each addend was supposed to be greater than or equal to sin60 so that when we add the three, we get the result. Yeah, it's a lousy solution, definitely wrong. So, basically i need help. nyehe...

I think we have discussed this type of problems several times on the forum. The idea is to write that $\frac{x}{1-x^2}\geq Ax^2$ where, of course, the constant $A$ has the marvelous property that we have equality exactly when $x^2+y^2+z^2=1$.

What's the constant here? Can you show?

$x \in [0,1]$ ${x^2 \over 1-x^2}\ge \frac{3\sqrt{3}}{2} x^3$ proof: $2 \ge 3\sqrt{3} x(1-x^2)$ $(x(1-x^2))'=1-3x^2=0$ just chek: $x={\sqrt{3} \over 3}$ sum it for $sin x,sin y , sinz$ and we are done...

_el_doopa wrote: $(x(1-x^2))'=1-3x^2=0$ just chek: $x={\sqrt{3} \over 3}$ and we are done... What do you mean?

indybar wrote: _el_doopa wrote: $(x(1-x^2))'=1-3x^2=0$ just chek: $x={\sqrt{3} \over 3}$ and we are done... What do you mean? have you learnt calculus? actually he means that the extremum is found when equating first derivative to 0. Second derivative test shows that the function attains its maximum when $x=\frac{\sqrt{3}}{3}$. So we only need to check x for that value to prove that inequality.

Second deriv. is $-6x$, right?

indybar wrote: Second deriv. is $-6x$, right? yes.

zhaoli wrote: Let $0 < \alpha, \beta, \gamma < \frac{\pi}{2}$ and $\sin^{3} \alpha + \sin^{3} \beta + \sin^3 \gamma = 1$. Prove that $\tan^{2} \alpha + \tan^{2} \beta + \tan^{2} \gamma \geq \frac{3 \sqrt{3}}{2}$. let $x=\sin \alpha$,and $y,z$ similar. then $x^3+y^3+z^3=1$ we should to prove:$\frac{x^2}{1-x^2}+\frac{y^2}{1-y^2}+\frac{z^2}{1-z^2} \geq \frac{3 \sqrt{3}}{2}$ which is equal to $\frac{x^3}{x(1-x^2)}+\frac{y^3}{y(1-y^2)}+\frac{z^3}{z(1-z^2)} \geq \frac{3 \sqrt{3}}{2}$ it is easy by $2x^2(1-x^2)^2 \le (\frac{2x^2+2-2x^2}3)^3$ but I think the equality can't be hold... so what is the minimum??

I think that the minimum is about 2.598 by using the Jensen's inequality.