# A Not So Simple Inequality in Three Variables

### Problem

### Solution 1

Assume, WLOG, $a+b+c=3\,$ such that $ab+bc+ca=3(1-t^2),\,$ $0\le t\lt 1.\,$ From here, $a^2+b^2+c^2=3(1+2t^2)\,$ and $a^3+b^3+c^3=27t^2+3p,\,$ where $p=abc.\,$ Thus, the required inequality is equivalent to

$\displaystyle\frac{27t^2}{p}+4\ge 4\left(\frac{1+2t^2}{1-t^2}\right)^2.$

According to Vo Quoc Ba Can, $p\le(1-t^2)(1+2t),\,$ implying that

$\displaystyle\frac{27t^2}{p}\ge \frac{27t^2}{(1-t)^2(1+2t)}.$

Hence, suffice it to show that

$\displaystyle\frac{27t^2}{(1-t)^2(1+2t)}\ge 4\left[\left(\frac{1+2t^2}{1-t^2}\right)^2-1\right]$

which is equivalent to

$\displaystyle\frac{27t^2}{(1-t)^2(1+2t)}\ge \frac{12t^2(2+t^2)}{(1-t^2)(1+t)^2}$

or, simplified,

$\displaystyle\frac{9}{1+2t}\ge \frac{4(2+t^2)}{(1+t)^2},$

i.e.,

$5t^2+2t+1\ge 8t^3,$

which is true for $t\in [0,1),\,$ because $8t^3-5t^2-2t-1=(t-1)(8t^2+3t+1).$

### Solution 2

Since the inequality is homogeneous, we may assume $p=\displaystyle\sum_{cycl}a=1.\,$ Let $r=abc\,$ and $\displaystyle\sum_{cycl}ab=\frac{1-q^2}{3},\,$ $0\le q\le 1.\,$ We have $\displaystyle\sum_{cycl}a^2=\frac{1+2q^2}{3}\,$ and $\displaystyle\sum_{cycl}a^3=q^2+3r.\,$

The required inequality is equivalent to

$\displaystyle\frac{q^2}{r}+3+1\ge\frac{\displaystyle\left(\frac{1+2q^2}{3}\right)^2}{\displaystyle\left(\frac{1-q^2}{3}\right)^2}.$

That's what we need to prove. Using Vo Quoc Ba Can's inequality,

$\displaystyle\frac{(1+q)^2(1-2q)}{27}\le r\le \frac{(1-q)^2(1+2q)}{27}$

we get

$\displaystyle\begin{align} LHS &= \frac{27q^2}{(1-q)^2(1+2q)}+4-4\left(\frac{1+2q^2}{1-q^2}\right)^2\\ &=\frac{3q^2(8q^2+3q+1)}{(1-q)(2q+1)(1+q)^2}\\ &\ge 0 \end{align}$

because $0\le q\le 1.$

Equality is achieved at $q=0,\,$ so that $p=1,\,$ $\displaystyle\sum_{cycl}ab=\frac{1}{3}\,$ and $\displaystyle r=\frac{1}{27},\,$ i.e., when $a=b=c=\displaystyle\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).$

### Acknowledgment

The problem, due to Richdad Phuc, has been kindly posted to the CutTheKnotMath facebook page, along with his solution, by Leo Giugiuc. Solution 2 is by Imad Zak.

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