Let $a$,$b$,$c$ be real numbers. Let $a$,$b$,$c$ be real numbers such that $a^{4}+b^{4}+c^{4}\leq27$. Prove the inequality:
$$\sum_{cyc}\frac{1}{\sqrt{a^{2}+1}}\geq \frac{3}{2}$$
My solution:
$\frac{1}{\sqrt{a^{2}+1}}\geq\frac{1}{2}\implies a^{2}\leq 3\implies a^{4}\leq 9$.
Similary : $b^{4}\leq9$ and $c^{4}\leq9$.
Then $a^{4}+b^{4}+c^{4}\leq27$ which is obviously true.
My solution is good ?

Let $x_a = \sqrt{a^2 + 1}$, and similarly for $x_b$ and $x_c$. Then by RMS-HM inequality on $x_a, x_b, x_c$,
$$\begin{align*}
\frac{3}{x_a^{-1}+x_b^{-1}+x_c^{-1}}
&\le \sqrt{\frac{x_a^2+x_b^2+x_c^2}3} \\
&= \sqrt{\frac{(a^2+1)+(b^2+1)+(c^2+1)}3}\\
&= \sqrt{\frac{a^2+b^2+c^2}3 + 1}\tag1\\
\end{align*}$$
By RMS-AM inequality on $a^2,b^2,c^2$,
$$\begin{align*}
\frac{a^2+b^2+c^2}3
&\le \sqrt{\frac{a^4+b^4+c^4}3}\\
&\le \sqrt{\frac{27}3}\\
&= 3\tag2\\
\end{align*}$$
Combining results $(1)$ and $(2)$,
$$\begin{align*}
\frac{3}{x_a^{-1}+x_b^{-1}+x_c^{-1}} &\le \sqrt{3+1}\\
\sum \frac{1}{\sqrt{a^2+1}} &\ge \frac{3}{2}\\
\end{align*}$$
