# Leo Giugiuc's Inequality with Constraint

### Solution

Let $p=a+b+c,\,$ $q=ab+bc+ca,\,$ $r=abc.\,$ We have

\displaystyle \begin{align} 2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)&=\frac{2(ab+bc+ca+2(a+b+c)+3)}{abc+ab+bc+ca+a+b+c+1}\\ &=\frac{2(q+2p+3)}{5+p}. \end{align}

After the substitution the required inequality becomes

\displaystyle \begin{align} \frac{2(q-2p+3)}{5+p}&\le q\,&\Leftrightarrow\\ (4-q)p+6&\le 3q. \end{align}

Now, since $q+r=4,\,$ implying $3\le q\le 4,\,$ and $q^2\ge 3pr\,$ we derive $\displaystyle 4\le q+\frac{q^2}{3p},\,$ i.e., $(4-q)p\le\displaystyle \frac{q^2}{3}.\,$ Suffice it to prove that $\displaystyle \frac{q^2}{3}+6\le 3q,\,$ which is equivalent to $(q-3)(q-6)\le 0\,$ which is true.

Equality is attained if $p=3,\,$ i.e., when $a=b=c=1.$

### Acknowledgment

Leo Giugiuc has kindly posted this problem of his at the CutTheKnotMath facebook page with a link to a solution by Richdad Phuc at the mathematical inequalities facebook group.