Let $a,b,c$ be non-negative reals and there are no $2$ of them are equal to $0$ simultaneously and satisfying: $a^2+b^2+c^2=1$. Prove that:$$\frac{{{a^3}}}{{\sqrt {{b^2} + {c^2}} }} + \frac{{{b^3}}}{{\sqrt {{c^2} + {a^2}} }} + \frac{{{c^3}}}{{\sqrt {{a^2} + {b^2}} }} \ge \frac{1}{{\sqrt {2(ab + bc + ca)} }}.$$
Attempt: I will be using $\sum$ to denote cyclic sums. Using the given condition $a^2+b^2+c^2=1$ and squaring both sides of the given inequality gives us this following equivalent inequality: $$\left[\sum \frac{a^3}{\sqrt{1-a^2}}\right]^2 \geq \frac{1}{2(\Sigma a b)} = \frac{a^2+b^2+c^2}{2(\Sigma ab)}$$ Applying Cauchy scharwz inequality on the LHS, we get $$\left[\sum \frac{a^3}{\sqrt{1-a^2}}\right]^2 \leq\left(\sum a^6\right)\left(\sum \frac{1}{1-a^2}\right)$$ Apart from this I also tried various manipulations but they didn't yield any useful result so I'm not adding those here. I am not looking for a solution, I want some instructive hints to proceed in this problem. Thanks.
Using Holder, we have $$\left(\sum_{\mathrm{cyc}}\frac{{{a^3}}}{{\sqrt {{b^2} + {c^2}} }}\right)^2\cdot \left(\sum_{\mathrm{cyc}} a(b^2 + c^2)\right) \cdot (a + b + c) \ge (a^2 + b^2 + c^2)^4.$$
It suffices to prove that $$\frac{(a^2 + b^2 + c^2)^4}{(a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2)(a + b + c)} \ge \frac{1}{2(ab + bc + ca)}$$ or $$\frac{a^2 + b^2 + c^2}{(a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2)(a + b + c)} \ge \frac{1}{2(ab + bc + ca)}$$ or (clearing the denominators) $$a^3b + b^3a + b^3c + bc^3 + c^3a + ca^3 - 2a^2b^2 - 2b^2c^2 - 2c^2a^2 \ge 0$$ which is true by AM-GM.
We are done.