# Quadratic Function for Solving Inequalities

### Problem

### Proof 1

If $xyz=0,\,$ there is nothing to prove. So, assume $xyz\ne 0.\,$ Denote $\displaystyle\frac{a}{x}=u,\,$ $\displaystyle\frac{b}{y}=v,\,$ $\displaystyle\frac{c}{z}=w.\,$ We need to prove that

$(u^2+3)(v^2+3)(w^2+3)\ge 4(u+v+w+1)^2.$

Consider the quadratic function

$f(u)=(u^2+3)(v^2+3)(w^2+3)-4(u+v+w+1)^2.$

Then $f'(u)=2u[(v^2+3)(w^2+3)-4]-8(v+w+1).\,$ Hence, $f\,$ admits a minimum value at $\displaystyle u=\frac{4(v+w+1)}{(v^2+3)(w^2+3)-4}.$

Let $\alpha=(v^2+3)(w^2+3)\ge 9\,$ and $\beta=v+w+1.\,$ We have

$\displaystyle f(u)\ge f\left(\frac{4\beta}{\alpha-4}\right)=\frac{[16\beta^2+3(\alpha-4)^2]\alpha}{(\alpha-4)^2}-\frac{4\beta^2\alpha^2}{(\alpha-4)^2}.$

Suffice it to show that $[16\beta^2+3(\alpha-4)^2]\alpha\ge 4\alpha^2\beta^2$ which is equivalent to $16\beta^2+3(\alpha-4)^2\ge 4\beta^2\alpha,\,$ or $(3\alpha-4\beta^2-12)(\alpha-4)\ge 0.\,$ But $\alpha -4\ge 0,\,$ thus, it remains to show that $3\alpha-4\beta^2-12\ge 0.\,$ Explicitly,

$(3v^2+5)w^2-8(v+1)w+5v^2-8v+11\ge 0.$

The discriminant of the quadratic function (in $w)\,$ on the left equals

$\Delta =-12(v-1)^2(5v^2+2v+13)\le 0$

so that the function never changes its sign and is not negative for all $w,\,$ thus proving the inequality.

### Proof 2

We start with

$(u^2+3)(v^2+3)(w^2+3)\ge 4(u+v+w+1)^2.$

and proceed without the recourse to derivatives. The inequality can be rewritten as

$(\alpha-4)w^2-8\beta w+3\alpha-4\beta^2\ge 0,$

where $\alpha=(u^2+3)(v^2+3)\gt 4\,$ and $\beta=u+v+1.\,$ The discriminant of the left-hand side

$\begin{align} \Delta &= (8\beta)^2-4(\alpha-4)(3\alpha-4\beta^2)\\ &= 4\alpha(-3\alpha+4\beta^2+12)\\ &= 4\alpha(-3u^2v^2-5u^2-5v^2+8uv+8u+8v-11)\\ &=-4\alpha[(uv-1)^2+(u-v)^2+4(u-1)^2+4(v-1)^2]\\ &\le 0, \end{align}$

with the same conclusion as in Proof 1.

### Acknowledgment

Leo Giugiuc has kindly communicated to me the problem, which he credits to Nguyen Viet Hung, with a solution of his (Proof 1). The problem has been posted at the mathematical inequalities facebook group where it was commented on by Mihai Dicu with another solution (Proof 2).

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny