<s>[INST] Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, then, using AM-GM inequality, we get 
$$
x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.
$$
Is it correct?
 [/INST] Let $f(t):=\tfrac{1+t^2}{1-t^2}$ so $f^{\prime\prime}(t)=\tfrac{4(1+3t^2)}{(1-t^2)^3}>0$. By Jensen's inequality, $f(a)+f(b)+f(c)\ge 3f(\tfrac13)=\frac{15}{4}$.
</s>