# a inequality posted on aops

Let $a,\,b,\,c$ are non-nagetive numbers such that $ab+bc+ca>0.$ Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+k \geqslant \frac{3}{2} + \frac{9k(a^3+b^3+c^3)}{(a+b+c)^3},$$where $k = \frac{13\sqrt{13}-35}{32}.$

Solution :at first use Nesbitt inequality whih follows that
(a/b+c)+(b/c+a)+(c/a+b)$\geq$ 3/2.
Then use AM of 3-rd power $\geq$ 3 rd power of A-M which follows that

$a^3+b^3+c^3$/$(a+b+c)^3$ $\geq$ 3/27 = 1/9 ,multypling into the both sides by 9 we get whole result $\geq$ 1.And multypling by k into the both sides we get the whole result $\geq$ k. then add two inequality and we will get the nesbitt inequality as:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+k \geqslant \frac{3}{2} + \frac{9k(a^3+b^3+c^3)}{(a+b+c)^3},$$\geq$ 3/2 + k( because we have prove the second inequality , that is $\geq$ 1.
Eliminate k from the both sides , and we will obtained the nesbitt inequality(john nesbitt , 1902 , England). so it is not true not only k = $k = \frac{13\sqrt{13}-35}{32}.$ , but also any positive values of k . So , we are done