<s>[INST] Proving $\sum\limits_{\rm cyc} 1/(a^2 -b+4) \geq3/4$ Suppose $a,b,c\in\mathbb R^+$ with $a+b+c=3.$ Prove that $$\frac{1}{a^2- b+4}+\frac{1}{b^2-c+4}+\frac{1}{c^2- a+4}\geqslant\frac{3}{4}.$$
I tried various approaches, but nothing seems to work. Whenever I give a lower bound using $C-S$ or $AM-GM$ it becomes very weak. And I am seeking solution that uses not so much computation.
 [/INST] By C-S
$$\sum_{cyc}\frac{1}{a^2-b+4}=\sum_{cyc}\frac{(a+5b+3c)^2}{(a+5b+3c)^2(a^2-b+4)}\geq\frac{81(a+b+c)^2}{\sum\limits_{cyc}(a+5b+3c)^2(a^2-b+4)}.$$
Thus, it's enough to prove that:
$$\frac{81(a+b+c)^2}{\sum\limits_{cyc}(a+5b+3c)^2(a^2-b+4)}\geq\frac{3}{4},$$ which is true and nice!
After homogenization we need to prove that
$$\sum_{cyc}(17a^4+23a^3b+23a^3c-78a^2b^2+15a^2bc)\geq0$$ or
$$17\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(40a^3b+40a^3c-78a^2b^2-2a^2bc)\geq0,$$ which is true by Schur and Muirhead. 
</s>