Is there a direct proof of the inequality $cyclic \sum\frac {a^2+1}{b+c}\ge 3$ Let $a,b,c>0$ reals.
Prove that
$$cyclic \sum\frac {a^2+1}{b+c}\ge 3$$
I proved it using Nesbitt inequality
$$cyclic \sum \frac {a}{b+c} \ge \frac {3}{2} $$
and the fact that
$$a+\frac {1}{a}\ge 2$$
But i would like to know if there is a straight proof without Nesbitt inequality.

Let $k=a+b+c$. Then
$$\frac{a^2+1}{b+c}=\frac{a^2-k^2+k^2+1}{k-a}=-a-k+\frac{k^2+1}{k-a}$$
So,
\begin{align*}
cyclic\sum\frac{a^2+1}{b+c}&=-a-b-c-3k+(k^2+1)\left(\frac{1}{k-a}+\frac{1}{k-b}+\frac{1}{k-c}\right)\\
&\ge -4k+3(k^2+1)\left(\frac{3}{(k-a)+(k-b)+(k-c)}\right)
\end{align*}
as $\textrm{A.M.}\ge\textrm{H.M.}$
\begin{align*}
-4k+3(k^2+1)\left(\frac{3}{(k-a)+(k-b)+(k-c)}\right)&=-4k+\frac{9(k^2+1)}{2k}\\
&=\frac{k}{2}+\frac{9}{2k}\\
&\ge3
\end{align*}
as $\textrm{A.M.}\ge\textrm{G.M.}$
