Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$$
Nesbitt's inequality is the following:
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2},$$ which follows from C-S: $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$ but this way does not help for the starting inequality.
There is a nice solution for the following inequality, which was in our test six months ago.
Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
By C-S $$\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{(b+c)^2}{ab+ac+2bc}.$$ Thus, it's enough to prove that $$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{(b+c)^2}{ab+ac+2bc}\geq\frac{3}{2}$$ or $$2(b+c)a^2-(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc)a+2(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq0.$$ Thus, it's enough to prove that $$16(b+c)(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ or $$16\sqrt[3]{4(b^3+c^3)^4}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ and since $$3(b+c)\sqrt[3]{4(b^3+c^3)}>4bc,$$ it remains to prove that $$4\sqrt[3]{2(b^3+c^3)^2}\geq3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc$$ and after using $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz),$$ where $$x^2+y^2+z^2-xy-xz-yz\neq0$$ we need to prove that $$128(b^3+c^3)^2-108(b+c)^3(b^3+c^3)+64b^3c^3+288(b+c)(b^3+c^3)bc\geq0.$$ Now, let $b^2+c^2=2kbc$. Hence, we need to prove that: $$128(2k+2)(2k-1)^2-108(2k+2)^2(2k-1)+64+288(2k+2)(2k-1)\geq0$$ or $$(k-1)^2(10k+11)\geq0.$$ Done!
But this way gives a wrong inequality again.