<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] Just another way is to note for $t \in (0, 1)$,
$$\frac{1+t^2}{1-t^2} \geqslant \frac{27t+11}{16} \iff (3t-1)^2(3t+5)\geqslant 0$$
Hence 
$$\sum_{cyc} \frac{1+a^2}{1-a^2}\geqslant \frac1{16}\sum_{cyc}(27a+11) = \frac{15}4 $$
</s>