as it turns out the function $ \displaystyle f(x) = \frac x{1-x^2}$ is convex on $(0,1)$ because \[ f''(x) = \frac { 4a(1-a^2)}{(1-a^2)^4} >0, \] so we can apply Jensen to obtain that \[ \sum \frac a{1-a^2} \geq 3 \frac { \frac s3}{ 1 - \frac {s^2}9 } \geq \frac {3\sqrt 3}2 \Leftrightarrow \] \[ \frac {s^2}9 + \frac 2{3\sqrt 3} s \geq 1 \quad (1) \] where $s=a+b+c$. But $(a+b+c)^2 \geq 3 (ab+bc+ca)=3$, so $s\geq 3$, so obviously (1) takes place. The equality can only follow when $a=b=c= \displaystyle \frac 1{\sqrt 3}$.

Another approach is following: $\frac{x}{1-x^2}\geq\frac{3\sqrt{3}}{2}\cdot x^2$.

When did the two of you become beginners? And yet another simple approach is to use the trigonometric substitution $a = \tan A, b = \tan B, c = \tan C, \mbox{ where } A + B + C = \pi/2$.

I am not guilty, Valentin did it first.

Myth wrote: I am not guilty, Valentin did it first. sorry, never gonna happen again

Could you bring the Singapore Selection Tests 2004? thank you very much... Moldova rulezzzz _________________________ "Our days are never coming back.... (System of a down -> Highway Song)"

I posted all 6 questions from the Singapore TST 2004, so they're lying in various threads around the forum. You can just do a search to find them.

Maybe you can gather up all the links like I did with the Romanian TST 2004 and open up a thread on the National Olympiads forum

By the way, Singapore TST is it just for beginners ??!!

Obviously it's not just for beginners, it just so happens that the 1st 2 questions were rather easy (as far as I see, so that everybody's marks would look a bit better). There were harder questions such as this and that.

It's a very old thread But I'm going to post a different (but easy to obtain) solution \[ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq\frac{3\sqrt{3}}{2}. \] $\iff \sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge 0$ Let $a\ge b\ge c$ $\implies 2a+\sqrt 3 a^2 - \sqrt 3 \ge 2b+\sqrt 3 b^2 - \sqrt 3 \ge 2c+\sqrt 3 c^2 - \sqrt 3$ And $1-a^2 \le 1-b^2 \le 1-c^2$ Thus by Tchebychef's inequality $3\sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge (\sum_{cyclic}2a+\sqrt 3 a^2 - \sqrt 3)(\sum_{cyclic} \frac{1}{1-a^2})$ Since $ab+bc+ca=1 \implies a+b+c \ge \sqrt 3$ and $a^2+b^2+c^2 \ge 1$ Combining the above two results $(\sum_{cyclic}2a+\sqrt 3 a^2 - \sqrt 3)(\sum_{cyclic} \frac{1}{1-a^2}) \ge 0$ $\implies \sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge 0$ QED

Easy. Just a Cauchy-Schwartz and it is equivalent to a+b+c>=sqrt{3}, which is trivial.

luofangxiang wrote: easy. see this: \[{\frac {\sqrt {{x}^{2} \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }}{{y}^{ 2}+{z}^{2}}}\geq 3/2\,{\frac {\sqrt {3}{x}^{2}}{{x}^{2}+{y}^{2}+{z}^{2}}}\] equality hold when $2x^2=y^2+z^2$

Similar, generalisated: sum a/(b^2+c^2)>=9/2sqrt(a^2+b^2+c^2).

Let $f(x)=\frac{x}{1-x^2}\text{ furthermore notice that }f''(x)=\frac{4x(1-x^2)}{(1-x^2)^4}>0\text{ for } x\in(0,1)$ thus the function is convex on this interval. Furthermore $\sum_{cyc}\frac{a}{1-a^2}=\sum_{cyc}f(a)\overset{\text{Jensen}}{\ge}3f\left(\frac{\sum_{cyc}a}{3}\right)=\frac{\sum_{cyc}a}{1-\frac{\left(\sum_{cyc}a\right)^2}{9}}$ Therefore the inequality boils down to $\frac{\sum_{cyc}a}{1-\frac{\left(\sum_{cyc}a\right)^2}{9}}\ge\frac{3\sqrt{3}}{2}\Longrightarrow2\sum_{cyc}a\ge3\sqrt{3}-\frac{\sqrt{3}\left(\sum_{cyc}a\right)^2}{3}\Longleftrightarrow\left(\sum_{cyc}a\right)\left(6+\sqrt{3}\sum_{cyc}a\right)\ge9\sqrt{3}$, which is clearly true as $\left(\sum_{cyc}a\right)^2\ge3\sum_{cyc}ab\Longrightarrow \sum_{cyc}a\ge\sqrt{3}$ $\blacksquare$.