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

Question

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 :)

Answer 1

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.

Answer 2

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!