# Marian Dinca's Refinement of Nesbitt's Inequality

### Proof

The first part,

\displaystyle\begin{align} \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}\\ &\ge\frac{3}{2}, \end{align}

was established in the derivation of Nesbitt's inequality from that of Bergström. The last part, viz.,

$\displaystyle\frac{3\sqrt{3(a^2+b^2+c^2)}}{2(a+b+c)}\ge\frac{3}{2},$

is equivalent to $\displaystyle\sqrt{\frac{a^2+b^2+c^2}{3}}\ge\frac{a+b+c}{3},$

i.e., the AM-QM inequality.

We need to prove the middle part:

$\displaystyle\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac{3\sqrt{3(a^2+b^2+c^2)}}{2(a+b+c)},$

i.e.,

$\displaystyle (a+b+c)^6\ge 27(ab+bc+ca)^2(a^2+b^2+c^2).$

Let $p=a+b=c,\;$ $q =ab+bc+ca.\;$ Then $a^2+b^2+c^2=p^2-2q\;$ and the inequality becomes

$\displaystyle p^6\ge 27q^2(p^2-2q).$

This is the same as $p^6-27q^3-27q^2(p^2-3q)\ge 0,\;$ which, in turn, reduces to

$\displaystyle (p^2-3q)(p^4+3p^2q-18q^2)\ge 0$

and, finally, to $(p^2-3q)^2(p^2+6q)\ge 0,\;$ which is true.

### Acknowledgment

I have borrowed the problem from the Imad Zak facebook group where it was posted by Dan Sitaru and commented on with the above proof by Diego Alvariz.

With an additional proof (by Leo Giugiuc) the result has been published in Romanian Mathematical Magazine.