Inequality Proof $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$

Question

Let $a,b,c\in \mathbb{R}^+$, and $a^2+b^2+c^2=1$, show that: $$ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$$

Answer 1

Here is a hint: it is enough to show for $x\in (0, 1)$, $$f(x) = \left(\frac{x}{1-x^2}-\frac{\sqrt3}2\right)-\frac{3\sqrt3}2\left(x^2-\frac13 \right) \geqslant 0$$

$$\iff \frac{x(\sqrt3x+2)(3x-\sqrt3)^2}{6(1-x^2)} \geqslant 0$$

Answer 2

The function $f(x)=\frac{1}{x(1-x^2)}$ takes its minimum at $x=\frac{1}{\sqrt 3}$ on $(0,1)$.

Thus $$\begin{eqnarray*}\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2} & = & a^2 f(a) + b^2 f(b) + c^2 f(c)\\ & \ge & (a^2+b^2+c^2)f\left(\frac{1}{\sqrt 3}\right)= \frac{3\sqrt 3}{2}.\end{eqnarray*}$$

Answer 3

By C-S and Schur we obtain: $$\sum_{cyc}\frac{a}{1-a^2}=\sum_{cyc}\frac{a^2}{ab^2+ac^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2b+a^2c)}\geq\frac{3(a+b+c)}{\sum\limits_{cyc}(a^2+ab)}.$$ Thus, it's enough to prove that $$\frac{a+b+c}{\sum\limits_{cyc}(a^2+ab)}\geq\frac{1}{2}\sqrt{\frac{3}{a^2+b^2+c^2}}.$$ Now, let $a^2+b^2+c^2=x(ab+ac+bc).$

Hence, $x\geq1$ and we need to prove that $$4(a+b+c)^2(a^2+b^2+c^2)\geq3\left(\sum_{cyc}(a^2+ab)\right)^2$$ or $$4(x+2)x\geq3(x+1)^2$$ or $$(x-1)(x+3)\geq0,$$ which is obvious.