Is there a direct proof of the inequality $cyclic \sum\frac {a^2+1}{b+c}\ge 3$

Question

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.

Answer 1

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 $\textrm{A.M.}\ge\textrm{G.M.}$

Answer 2

Hint: Using CS and then AM-GM, $$\sum \frac{a^2}{b+c} +\sum \frac1{b+c}\geqslant \frac{a+b+c}2+\frac9{2(a+b+c)}\geqslant 3$$

Answer 3

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{a^2+1}{b+c}=\sum_{cyc}\frac{a^2}{b+c}+\sum_{cyc}\frac{1}{b+c}\geq2\sqrt{\sum_{cyc}\frac{a^2}{b+c}\sum_{cyc}\frac{1}{b+c}}\geq$$ $$\geq2\sqrt{\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+c)}\sum_{cyc}\frac{1}{b+c}}=\sqrt{2(a+b+c)\sum_{cyc}\frac{1}{b+c}}=$$ $$=\sqrt{\sum\limits_{cyc}(b+c)\sum_{cyc}\frac{1}{b+c}}\geq\sqrt{(1+1+1)^2}=3.$$