Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello I just wondering if it's possible to prove this inequality: there are positive, various $a,b,c$ and

$ \frac{3a-b}{3} \ge x \ge \frac{3(a^2-b^2)}{3a+b}$

$ \frac{3b-c}{3} \ge y \ge \frac{3(b^2-c^2)}{3b+c}$

$ \frac{3c-a}{3} \ge z \ge \frac{3(c^2-a^2)}{3c+a}$

I want to prove that $x+y \ge z$ when we add we get

$\frac{3a+2b-c}{3} \ge x+y \ge \frac {3(a-b)(a+b)}{3 a+b}+\frac {3(b-c)(b+c)}{3 b+c} $ but further I don't know how to compare it with $z$

share|cite|improve this question


To prove that $x+y\ge z$ we need to prove that minimum value of $x$ plus minimum value of $y$ is $\ge$ maximum value of $z$.

So you have:

$x=\frac{3(a^2-b^2)}{3a+b}; y=\frac{3(b^2-c^2)}{3b+c}; z=\frac{3c-a}{3}$

Just prove that $\frac{3(a^2-b^2)}{3a+b}+\frac{3(b^2-c^2)}{3b+c}\ge\frac{3c-a}{3}$

share|cite|improve this answer
ok, thank you :) – Marco Nov 14 '13 at 16:30
let me know how it's going :) – Gintas K Nov 14 '13 at 16:37
@Marco I solved this and got:… take a look :) So this equality is correct only if a,b and c meets some conditions :) – Gintas K Nov 14 '13 at 16:49
looks a bit complicated, and only the second condition satisfy the assumptions – Marco Nov 14 '13 at 16:56
So we proved that this equality is incorrect with the given limits of x,y,z :) – Gintas K Nov 14 '13 at 16:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.