Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$

Question

I was given some exercises from Math olympiads, and I am stuck with the one below, which seems soluble, yet I can't come up with something that works.

Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$.

I tried AM-GM-HM, forming squares, but nothing appears to work.

If you have any idea on the source of this problem, that would be also welcomed.

First i tried using AM-GM for a+b+c, and then tried to prove that the remaining is >= 1. After trying, I just tested it using a=1, b=2, c=1/2, and realized that this is not true in general.

Next I tried dividing with abc, did some calculations and ended up with $\sum\limits_{cyc}(a−1)(b+c)^2\geq0$ , which also is not true. Similarly, I got $\sum\limits_{cyc}\frac{(a+b)^2}{ab}\geq\sum\limits_{cyc}2\left(\frac{1}{a}+\frac{1}{b}\right)$ , which also is not true. Next I found an intermediate result in the web, and I was left with proving AM*HM > 1, which also is not true in this case, apparently.

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$3u+\frac{w^3}{v^2}\geq 4w,$$ which is true by AM-GM: $$3u+\frac{w^3}{v^2}\geq4\sqrt[4]{\frac{u^3w^3}{v^2}}\geq4\sqrt[4]{\frac{wv^2w^3}{v^2}}=4w.$$ The following inequality is also true.

Let $a$, $b$ and $c$ be positive numbers such that $abc=1.$ Prove that: $$a+b+c+\frac{8}{ab+ac+bc}\geq\frac{17}{3}.$$

Answer 2

Elaborating on my hint, show that

1. $(a+b+c)^2 \geq 3(ab+bc+ca)$
2. $a+b+c \geq 3$
3. let $ t = a+b+c$, then $LHS \geq t + \frac{ 9}{t^2}$.
4. Finally, show that for $ t \geq 3$, $ t + \frac{9}{t^2 } \geq 4$.
   • Lots of ways to do this.