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?
 A: 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, $f(a)+f(b)+f(c)\ge 3f(\tfrac13)=\frac{15}{4}$.
A: By C-S
$$a^2+b^2+c^2=\frac{1}{3}(1+1+1)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2=\frac{1}{3}.$$
Thus, by C-S again we obtain:
$$\sum_{cyc}\frac{1+a^2}{1-a^2}=\sum_{cyc}\left(1+\frac{2a^2}{1-a^2}\right)\geq3+\frac{2(a+b+c)^2}{\sum\limits_{cyc}(1-a^2)}=$$
$$=3+\frac{2}{3-(a^2+b^2+c^2)}\geq3+\frac{2}{3-\frac{1}{3}}=\frac{15}{4}.$$
A: Your way gives a right inequality!
Indeed, by AM-GM
$$\sum_{cyc}\frac{1+a^2}{1-a^2}\geq3\sqrt[3]{\prod_{cyc}\frac{1+a^2}{1-a^2}}.$$
Thus, it's enough to prove that
$$64\prod_{cyc}(1+a^2)\geq125\prod_{cyc}(1-a^2)$$ or
$$\sum_{cyc}\left(\ln(1+a^2)-\ln(1-a^2)+2\ln2-\ln5\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(1+a^2)-\ln(1-a^2)+2\ln2-\ln5-\frac{27}{20}\left(a-\frac{1}{3}\right)\right).$$
Now, let $$f(a)=\ln(1+a^2)-\ln(1-a^2)+2\ln2-\ln5-\frac{27}{20}\left(a-\frac{1}{3}\right).$$
Thus, $$f'(a)=\frac{2a}{1+a^2}-\frac{-2a}{1-a^2}-\frac{27}{20}=\frac{(3a-1)(9a^3+3a^2+a+27)}{20(1-a^4)},$$
which gives $$a_{min}=\frac{1}{3}$$ and since $$f\left(\frac{1}{3}\right)=0,$$ we are done!
A: 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 
$$\sum_{cyc} \frac{1+a^2}{1-a^2}\geqslant \frac1{16}\sum_{cyc}(27a+11) = \frac{15}4 $$
A: No it’s wrong. Hint : use the CS inequality but the “ Angel “ form. 
A: Hint: You can solve by showing $${1\over 1-x^2} \geq {27\over 32}(x+1)$$ for $x\in (0,1)$.
