An Inequality with Constraints II

Leonard Giugiuc, Dan Sitaru 3 December, 2015

For real numbers $a,b,c$ such that $3\gt a\ge 1\ge b, c\ge 0,\;$ and $a+b+c=3,$

$\displaystyle abc+\frac{2}{ab+bc+ca}\ge\frac{5}{a^2+b^2+c^2}.$

Proof

Denote $abc=p$ and $ab+bc+ca=q.\;$ Introduce the polynomial $f(x)=(x-a)(x-b)(x-c).$ From $a\ge 1\ge b, c,$ $f(1)\le 0.$ On the other hand, $f(x)=x^3-3x^2+qx-p$ so that $f(1)=-2+q-p.$ Thus $p\ge q-2.$ Let's consider two cases:

Case 1: $2\le q\le 3.$

We have

$\displaystyle abc+\frac{2}{ab+bc+ca}=p+\frac{2}{q}\ge q-2+\frac{2}{q}.$

Since $a^2+b^2+c^2=9-2q,$ suffice it to show that $\displaystyle q-2+\frac{2}{q}\ge\frac{5}{9-2q}.$ The latter is in turn equivalent to $(3-q)(q-2)(2q-3)\ge 0,$ which is obviously true in this case.

Case 2: $0\lt q\lt 2.$

In this case, $\displaystyle q-2+\frac{2}{q}\ge\frac{2}{q}\gt 1\;$ and $\displaystyle\frac{5}{9-2q}\lt 1,$

which completes the proof.

(This was posted at the CutTheKnotMath facebook page.)