# For $x, y > 0$, can you show that $\frac{x(2x-y)}{y(2z + x)} + \frac{y(2y-z)}{z(2x+y)}+\frac{z(2z-x)}{x(2y+z)}\geqslant 1$

I tried going for a common denominator but then it turned the whole inequality into a big muddle... I also tried multiplying the brackets out but to not much avail...

I also browsed through some known inequalities such as Cauchy and AM-GM (I only know the simpler ones) but I couldn't find any known inequalities that I could use in this problem.

Also, I tried to look for other problems on this site with similar inequalities, but I couldn't find much relevant inequalities.

I'm legit stuck.

If $z=0$, then there is a divisibility by $0$ and the world explodes.

Here is the question again:

For $x, y > 0$, can you show that

$$\frac{x(2x-y)}{y(2z + x)} + \frac{y(2y-z)}{z(2x+y)}+\frac{z(2z-x)}{x(2y+z)}\geqslant 1$$

Multiplied out:

$$\frac{2x^{2}-xy}{2yz+xy} + \frac{2y^{2}-yz}{2xz+zy} + \frac{2z^{2}-xz}{2xy+xz}\geqslant 1$$

Common Denominator would be $xyz(2z+x)(2x+y)(2y+z)$

Any contribution will be deeply appreciated.

Please don't mark this one as 'unconstructive' or 'duplicate' because I really need an answer for an assignment. Help me out please?

Thanks a lot :)

• Posts like these get shut down because you havent shown any effort to try and solve the problem yourself. – user45878 Feb 24 '13 at 19:52
• You seemed okay with the question being closed yesterday. Also, this site is community-run; as you get more reputation points, you get more abilities to help out the moderators on the site. Anyone with more than 3000 points can vote to close a question; the people who closed your original question were not "admins". – Zev Chonoles Feb 24 '13 at 19:53
• @mathsnoob We are here to help you understand mathematics and answer questions about it. But that is not the same as being here to do your homework. As it stands, it looks like you are just asking us to solve this exercise for you. If you were to edit the post to include your thoughts on the question and what you've tried so far, people would be much more inclined to help you. – Alex Becker Feb 24 '13 at 19:59
• @mathsnoob Look the best way to get an answer on this site is to do this. At the top of your question, write the statement, i.e. "Prove that [insert inequality] is true." Then below that write out your detailed scratch work, like "Begin with this assumption. Then this is true then this is true blah blah." Your attempt at the proof. Then, somewhere you'll get stuck. At that point, ask a specific question, like "why doesn't this work?" or "can I get a hint on this next step?" rather than "OK now solve this problem for me." The reason this will help you is that it will show us you cared – Alexander Gruber Feb 24 '13 at 20:16
• @mathsnoob If somebody solves it for you they're going to have to take AAAGGES to type it too. I don't see why you think we're some kind of on-demand solutions manual. It's not like we get paid for this or anything. – Alexander Gruber Feb 24 '13 at 21:46

## 2 Answers

I believe that the deadline of the assignment is probably over, so let me post an idea.

Note that

$$\sum \frac{x^2}{2yz + xy} \ge \frac{(\sum x)^2}{3(xy+yz+zx)} \ge 1$$

by Cauchy-Schwarz. So it suffices to show that $\displaystyle \sum \frac{x^2-xy}{2yz+xy} \ge 0$. Note that this is equivalent to

\begin{align} &\sum_{cyc} (x^2-xy) \left( \frac{1}{2yz+xy} - \frac{1}{x^2+xy+xz} \right) \ge 0 \\ &\Leftrightarrow \sum (x^2-xy) \frac{x^2 - xz + 2xz - 2yz}{(2yz+xy)(x^2+xy+xz)} \ge 0 \\ &\Leftrightarrow \sum \frac{x^2(x-y)(x-z)}{(2yz+xy)(x^2+xy+xz)} + \sum 2xz\frac{(x-y)^2}{(2yz+xy)(x^2+xy+xz)} \ge 0 \end{align}

It suffices to show that the first sum is non-negative. We further simplify it,

$$\sum \frac{x^2(x-y)(x-z)}{(2yz+xy)(x^2+xy+xz)} = \frac{1}{x+y+z} \sum \frac{x(x-y)(x-z)}{(2yz+xy)}$$

and it now suffices to show that the inner sum is non-negative. Since the sum is cyclic, WLOG assume that $x$ is the largest.

If $x \ge y \ge z$, note that among the three summands, only the $y(y-z)(y-x)$ term is negative. But note that when we consider it with the $x(x-y)(x-z)$ term together, it is \begin{align} &\frac{x(x-y)(x-z)}{(2yz+xy)} + \frac{y(y-z)(y-x)}{(2zx+yz)} \\ &\ge \frac{x(x-y)(y-z)}{(2yz+xy)} + \frac{y(y-z)(y-x)}{(2zx+yz)} \\ &= (x-y)(y-z) \left(\frac{x}{2yz+xy} - \frac{y}{2zx+yz}\right) \\ \end{align} The denominator of the bracket is $2x^2z + xyz - 2y^2z - xyz = 2z(x^2-y^2) \ge 0$, so we are done for this case.

If $x \ge z \ge y$, note that among the three summands, only the $z(z-x)(z-y)$ term is negative. But note that when we consider it with the $x(x-y)(x-z)$ term together, it is \begin{align} &\frac{x(x-y)(x-z)}{(2yz+xy)} + \frac{z(z-x)(z-y)}{(2xy+zx)} \\ &\ge \frac{x(z-y)(x-z)}{(2yz+xy)} + \frac{z(z-x)(z-y)}{(2xy+zx)} \\ &= (z-y)(x-z) \left(\frac{x}{2yz+xy} - \frac{z}{2xy+zx}\right) \\ \end{align} The denominator of the bracket is $2x^2y + x^2z - 2yz^2 - xyz = 2y(x^2-z^2) + xz(z-y) \ge 0$, so we are also done for this case.

• Thanks! Yes, the deadline is over :) – mathsnoob Mar 2 '13 at 6:46

A full expanding gives: $$\sum_{cyc}(x^4z^2+2x^4yz-x^3y^2z-x^3z^2y-x^2y^2z^2)\geq0$$ or $$xyz\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}(x^4z^2-x^2y^2z^2)+xyz\sum_{cyc}(x^3-xyz)\geq0,$$ which is Muirhead and AM-GM.

Done!