# Stronger than Nessbit

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{3\sqrt[3]{a^2b^2c^2}}{2(a^2+b^2+c^2)}\geq2\tag{1}$$

we know this Nessbit inequality $$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{(a+b+c)^2}{2(ab+bc+ac)}\ge\dfrac{3}{2}$$ and use AM-GM inequality $$a^2+b^2+c^2\ge 3\sqrt[3]{a^2b^2c^2}\Longrightarrow \dfrac{3\sqrt[3]{a^2b^2c^2}}{2(a^2+b^2+c^2)}\le\dfrac{1}{2}$$ so $(1)$ is stonger than Nessbit inequality

Now so How to prove $(1)$

• Hi, from your inequalities and homogeneity of $(1)$ we take $abc=1$ remains to prove that $$\frac{a^2+b^2+c^2}{2(ab+bc+ca)}+\frac{3}{2(a^2+b^2+c^2)}\ge 1$$ ? – Toni Mhax May 3 '18 at 4:11

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $$\sum_{cyc}\frac{a}{b+c}+\frac{3w^2}{2(9u^2-6v^2)}\geq2$$ or $$\frac{\sum\limits_{cyc}(a(a+b)(a+c)}{\prod\limits_{cyc}(a+b)}+\frac{w^2}{2(3u^2-2v^2)}\geq2$$ or $$\frac{\sum\limits_{cyc}(a^3+a^2b+a^2c+abc)}{9uv^2-w^3}+\frac{w^2}{2(3u^2-2v^2)}\geq2$$ or $$\frac{27u^3-27uv^2+3w^3+9uv^2-3w^3+3w^3}{9uv^2-w^3}+\frac{w^2}{2(3u^2-2v^2)}\geq2$$ or $$\frac{27u^3-18uv^2+3w^3}{9uv^2-w^3}+\frac{w^2}{2(3u^2-2v^2)}\geq2$$ or $f(w^3)\geq0,$ where $$f(w^3)=-w^5+30u^2w^3-20v^2w^3+9uv^2w^2+162u^5-324u^3v^2+144uv^4.$$ We see that $f$ is a concave function.

But, the concave function gets a minimal value for an extreme value of $w^3$, which happens in the following cases.

1. $w^3=0$.

Let $c=0$. We need to prove that $\frac{a}{b}+\frac{b}{a}\geq2,$ which is true by AM-GM;

1. Two variables they are equal.

Since our inequality is symmetric and homogeneous, it's enough to assume that $b=c=1$.

Let $a=x^3$.

Thus, we need to prove that $$\frac{x^3}{2}+\frac{2}{x^3+1}+\frac{3x^2}{2(x^6+2)}\geq2$$ or $$x^2(x-1)^2(x^8+2x^7+3x^6+x^5-x^4-3x^3-3x^2+3)\geq0,$$ which is easy.