# Prove when $abc=1$: $\frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$

Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$: $$\frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$

My thoughts were turning the right hand side to $abc$ as $abc=1$ however I think this will make the proving even more harder. I also attempted applying the Cauchy-Schwarz inequality and Hölder's inequality but to no avail.

Could someone please show me how to prove it using the inequalities above or another method.

If $a,b,c> 0$ and $abc=1$, then by Cauchy-Schwarz: $$\sum_{\text{cyc}}\frac{a}{2+bc}=\sum_{\text{cyc}}\frac{a^2}{2a+abc}=\sum_{\text{cyc}}\frac{a^2}{2a+1}\ge \frac{(a+b+c)^2}{2(a+b+c)+3}$$

And by AM-GM $a+b+c\ge 3\sqrt[3]{abc}=3$, so: $$(a+b+c)^2-2(a+b+c)-3=((a+b+c)-3)((a+b+c)+1)\ge 0$$

Equality holds if and only if $a=b=c=1$.

• Actually, $(a+b+c)^2\ge 3(a+b+c)=2(a+b+c)+(a+b+c)\ge 2(a+b+c)+3$. – Colescu Apr 30 '16 at 7:30
• @YuxiaoXie His method works too... – S.C.B. Apr 30 '16 at 7:31
• @MXYMXY Indeed, but maybe just a bit more "tedious". I was just providing another proof anyway. – Colescu Apr 30 '16 at 7:37

Let $p,q,r \in \mathbb{R}_{>0}$ such that $(p^3,q^3,r^3)=(a,b,c)$.

Then $(pqr)^3 = 1$ and hence $pqr = 1$.

Thus $\sum_{cyc} \frac{a}{2+bc} \ge 1$ iff $\sum_{cyc} \frac{p^3 (pqr)}{2 (pqr)^2 + q^3 r^3} \ge 1$ iff $\sum_{cyc} \frac{p^4}{2 p^2 q r + q^2 r^2} \ge 1$.

This technique is called homogenization, and many homogenous inequalities can be solved by expanding and using AM-GM and Schur's. I didn't try for this one though.

Note that $$\sum_{cyc}\frac{a}{bc+2}=\sum_{cyc}\frac{a^2}{2a+abc}=\sum_{cyc}\frac{a^2}{2a+1} \ge \frac{(a+b+c)^2}{2a+2b+2c+3}$$

Now one can use that $$\sum_{cyc}a \ge 3\sqrt[3]{abc}=3$$This establishes that $$(a+b+c-1)^2 \ge 4 \Leftrightarrow (a+b+c)^2 \ge 2(a+b+c)+3$$Thus, our proof is done, with equality at $a=b=c$.

• Please would you mind elaborating a bit on the last step in the first line? Is that a popular inequality? Does it hold always? (sum of fractions is greater than sum of numerators divided by sum of denominators?) – Aritra Das Apr 30 '16 at 7:33
• Actually, $(a+b+c)^2\ge 3(a+b+c)=2(a+b+c)+(a+b+c)\ge 2(a+b+c)+3$. I think it's easier to understand because your way of proving $(\sum_{cyc} a)^2\ge 2(\sum_{cyc} a)+3$ seems to come out of the blue (though brilliant)! – Colescu Apr 30 '16 at 7:33
• @AritraDas It's Cauchy-Schwarz inequality in Engel form. You'll see why it's true once you multiply both sides by the denominator and apply Cauchy-Schwarz inequality. – Colescu Apr 30 '16 at 7:33