# Olympiad inequality $\frac{a}{2a + b} + \frac{b}{2b + c} + \frac{c}{2c + a} \leq 1$.

I want to prove that for $a, b, c > 0$ we have $\frac{a}{2a + b} + \frac{b}{2b + c} + \frac{c}{2c + a} \leq 1$.

My approach: I know that each of the individual terms is lesser than $\frac{1}{2}$ because of it's form. I am familiar with the Cauchy-schwarz inequality and the AM-GM-HM inequality. I tried using AM-GM but could not get anywhere because of the way the inequality is structured. Similarly I tried using Cauchy-schwarz as well.

I just need some intuition/hints on how to actually reduce this to a feasible form to solve and not the actual answer because it does not provide me with the necessary intuition which I need as I am a beginner when it comes inequalities. If I can't use these 2 inequalities, is this problem best tackled with some other inequality?

• Make common denominator on left hand side and use AM-GM to prove numerator$\leq$ denominator – user175968 Jul 23 '15 at 7:49
• Maybe it is brutal, but eliminate the denominators gives you the equivalent inequality $$3abc \leq a^2 c + ab^2 + bc^2,$$ which can be shown by AM-GM inequality. – Zhanxiong Jul 23 '15 at 7:53
• Ah that form is way easy to apply. I thought there would be a more clever way instead of expanding the denominator. But since I am a beginner I will develop tricks with more practice I think. Thanks for the hint! – iart Jul 23 '15 at 7:57
• – Martin Sleziak Apr 20 '17 at 1:47

hint:

$1-\dfrac{2a}{2a+b}=\dfrac{b}{2a+b}=\dfrac{b^2}{2ab+b^2}$

• Oh I see applying cauchy schwarz, we get cyclic sum($\frac{b^2}{2ab + b^2}) \geq \frac{(a + b + c)^2}{(a + b + c)^2} = 1$. Therefore we see that CyclicSum($1 - \frac{2a}{2a + b}$) $\geq$ 1. Therefore CyclicSum($\frac{a}{2a + b}$) $\leq 1$ which is the proof for the inequality. Very elegant. – iart Jul 23 '15 at 8:03
• Now the question is: What was the intuition behind that manipulation you did in the hint @chenbai? I am a beginner so looking to learn tricks. – iart Jul 23 '15 at 8:05
• if you see some examples posted before in SE, you will find this is a trick to change the direction of the inequality that make one can apply the well known inequality. – chenbai Jul 23 '15 at 9:06

Hint:

Set $x = \frac{b}{a}, \, y = \frac{c}{b}, z = \frac{a}{c}$. The LHS becomes $$\sum_{cyc.}\frac{1}{2 + x}$$ with $xyz = 1.$

• Ah I like this trick. I understand the intuition behind the hint also. Do you know what the intuition behind the hint for the previous answer is? – iart Jul 23 '15 at 8:17
• Would you mind elaborating on the hint a bit? It seems to me easier to prove with the constraint unwound than under it. So I would very much like to see the trick. – Hans Sep 3 '17 at 4:19
• @iart or @ corindo Could you please explain what is the intuition behind the answer? I don't see it... – Ovi Jul 2 '18 at 14:02