Answer is $\frac{1}{8}$, just use $\frac{x^3}{1-x^2} \geq \frac{13x-3}{32}$.

The Jensen work too.

Another one using Holder: $E(a,b,c)\cdot [(1-a)+(1-b)+(1-c)]\cdot [(1+a)+(1+b)+(1+c)]\geq(a+b+c)^3$ $\Leftrightarrow E(a,b,c)\geq \dfrac{1}{8}$ Equality holds if $a=b=c=\dfrac{1}{3}$

Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1.$ Prove that$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac98$$$$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}\geq \frac38$$$$\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}\geq \frac18$$

Positive real numbers $a$, $b$, $c$ satisfy $a^2+b^2+c=1.$ Prove that$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq 2$$$$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}\geq1$$$$\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}\geq \frac12$$

sqing wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1.$ Prove that$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac98$$$$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}\geq \frac38$$$$\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}\geq \frac18$$ $$\sum \frac{a}{1-a^2} = \sum \frac{a^2}{a-a^3}\ge \sum \frac{1}{3-\sum a^3} \ge \frac{9}{8}$$since Titu's lemma and $(1+1+1)(1+1+1)(a^3+b^3+c^3) \ge (a+b+c)^3$ $$\sum \frac{a^2}{1-a^2}\ge \frac{1}{3- \sum a^2} \ge \frac{3}{8}$$since Titu's lemma and $(1+1+1)(a^2+b^2+c^2) \ge (a+b+c)^2$ $$\sum \frac{a^3}{1-a^2} = \sum \frac{a^4}{a-a^3} \ge \frac{(a^2+b^2+c^2)^2}{1- \sum a^3} \ge \frac{1}{8}$$since Titu's lemma and previous results.

Positive real numbers $a$, $b$, $c$ satisfy $a^2+b^2+c=1.$ Prove that$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\leq 2\sqrt 2$$$$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}\leq2$$$$\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}\leq\sqrt 2$$

augustin_p wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1$. Find the smallest possible value of $E(a,b,c)=\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}$. One can kill this also via rearrangement. Let $a\ge b\ge c$ w.l.o.g. Then $(1-a^2)^{-1}\ge (1-b^2)^{-1}\ge (1-c^2)^{-1}$; and thus \[ E(a,b,c)\ge \frac13\left(a^3+b^3+c^3\right)\left(\frac{1}{1-a^2}+\frac{1}{1-b^2}+\frac{1}{1-c^2}\right). \]Now establish for $S_n:=a^n+b^n+c^n$, $S_3\ge 1/9$ and $S_2\ge 1/3$ via power mean and QM-AM inequalities, respectively; and apply the AM-HM to the second term above to conclude (and avoid fancy Holder machinery )

Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1.$ Prove that$$\frac{a^k}{1-a^2}+\frac{b^k}{1-b^2}+\frac{c^k}{1-c^2}\geq \frac{3^{3-k}}{8}$$Where $k\geq 1.$ sqing wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1.$ Prove that$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac98$$ sqing wrote: Positive real numbers $a$, $b$, $c$ satisfy $a^2+b^2+c=1.$ Prove that$$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}\geq1$$

We claim that the minimum possible answer is $\boxed{\frac{1}{8}}$ By Jensen, let $f(x)=\frac{x^3}{1-x^2}$ After some boring computations we found out that $f$ is convex over $[0,1)$ Applying jensen's inequality we get, $\frac{f(a)+f(b)+f(c)}{3} \geq f(\frac{a+b+c}{3}) \Rightarrow f(a)+f(b)+f(c) \geq 3f(\frac{1}{3})=\frac{1}{8}$ Equality holds when $a=b=c=\frac{1}{3}$

By applying Hölder's inequality we get $(\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2})((1-a^2)+(1-b^2)+(1-c^2))(1+1+1) \geq (a+b+c)^3=1$ and also $3(a^2+b^2+c^2) \geq (a+b+c)^2=1$ so $ \frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2} \geq \frac{1}{9-3(a^2+b^2+c^2)} \geq \frac{1}{8}$

How about simply \[ \frac{a^3}{(1 - a)(1 + a)} + \frac{1 - a}{16} + \frac{1 + a}{32} \geq \frac{3a}{8} \]

augustin_p wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1$. Find the smallest possible value of $E(a,b,c)=\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}$. We have by Titu's Lemma $E(a, b, c) = \frac{a^4}{a-a^3} + \frac{b^4}{b-b^3} + \frac{c^4}{c-c^3} \geqslant \frac{(a^2+b^2+c^2)^2}{1-(a^3+b^3+c^3)}$. By Power Mean $a^2+b^2+c^2 \geqslant 1/3$ and $a^3+b^3+c^3 \geqslant 1/9$. This finishes since we get $E(a, b, c) \geqslant \frac{1/9}{1-1/9} = 1/8$.

sqing wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1.$ Prove that$$\frac{a^k}{1-a^2}+\frac{b^k}{1-b^2}+\frac{c^k}{1-c^2}\geq \frac{3^{3-k}}{8}$$Where $k\geq 1.$ sqing wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1.$ Prove that$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac98$$ sqing wrote: Positive real numbers $a$, $b$, $c$ satisfy $a^2+b^2+c=1.$ Prove that$$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}\geq1$$ I think my approach above can be adapted to solve this problem by breaking into cases by parity of $k$.

augustin_p wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1$. Find the smallest possible value of $$E(a,b,c)=\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}.$$ $E(a,b,c)>=\frac{(a+b+c)^3}{9-3(a^2+b^2+c^2}>=\frac{1}{8}$ The last one is true because $-1>=-3(a^2+b^2+c^2)$