# Olympiad Inequality - A not-so-Open Problem

I saw this problem online some time ago, and I have been trying to solve this inequality:

$x,y,z >0$, prove $\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$

This question was asked online years ago and no one has proved it with an "elegant" way. I decided to share it here to all who haven't seen this problem yet, as I find it interesting and exciting while solving the problem (although I haven't found an elegant proof) XD

Have fun!

Upon typing the title I remembered a quote I saw in Evan Chen's book :D :

Graders received some elegant solutions, some not-so-elegant solutions, and some so-not-elegant solutions. --MOP 2012

Note by Hua Zhi Vee
1 year, 4 months ago

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There is a solution that involves using symmetric polynomials, but you will need a good computer to do this.

- 1 year, 4 months ago

This problem can be solved using just pen and paper.

This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.

- 1 year, 4 months ago

Of course you can, but a proper proof would require symmetric polynomials, for which simplifying them (in this case) would be a right piece of work for a computer, let alone a human being. Fortunately, I have a very good PC that can help me out here; many others on this website would struggle to do so.

- 1 year, 4 months ago

Ok. I was wondering if it could be solved using inequalities/theorems like these.

- 1 year, 4 months ago

- 2 months ago

Colorectal cancer (CRC) is one of the most common and deadly cancers in the world. CRC occurs due to the accumulation of genetic interactions and alterations. At present, the molecular research progress of CRC is still slow. It has been reported that there is a close relationship between GGCT expression and CRC cells.

https://www.creative-biogene.com/genesearch/C7orf24.html

- 6 months ago

Hey,

$\displaystyle \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \geq \frac{x+y+z}{13} \geq \frac{3(xyz)^\frac{1}{3} }{13}$

So, can we prove $\displaystyle \ \ \ \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \geq \frac{3(xyz)^\frac{1}{3} }{13}$

Cause if we can, Then it is quite easy.... IN THE CASE WHEN $\ \ x,y,z \geq 0$

- 2 months ago

SOLUTION

$\displaystyle \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \geq \frac{x+y+z}{13} \geq \frac{3(xyz)^\frac{1}{3} }{13}$

So We can say $\displaystyle \ \ \ \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \times 13(x^3 + y^3 +z^3) \geq (x^2 + y^2 + z^2)^2 \ \ \ \$ By Cauchy-Schwarz Inequality more precisely Titu's Lemma

Also we can see that $\displaystyle \ \ x^2 + y^2 + z^2 \geq xy + yz + zx$

Thus, $\displaystyle \ \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \times (x^3 + y^3 +z^3) \geq \frac {(xy + yz + zx)^2}{13}$

And, $\displaystyle \ \ \frac {(xy + yz + zx)^2}{13} \geq \frac {9(xyz)^ \frac {4}{3} }{13} \ \$ By AM-GM-HM Inequality

Finally, $\displaystyle \ \ \frac {(xy + yz + zx)^2}{13} \geq \frac {3(xyz)^ \frac {1}{3} }{13} \times 3xyz \ \$

So$\displaystyle \ \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \times (x^3 + y^3 +z^3) \geq \frac {3(xyz)^ \frac {1}{3} }{13} \times 3xyz$........................1

Also, $\displaystyle \ \ x^3 + y^3 +z^3 \geq 3xyz$.....................2

Now, as $\displaystyle \ \ x,y,z \geq 0 \ \$ So divide inequality 1 by inequality 2 ,

Hence the final inequality becomes $\displaystyle \ \ \ \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \geq \frac{3(xyz)^\frac{1}{3} }{13}$

If it wasn't metioned that$\displaystyle \ \ x,y,z \geq 0 \ \$ we couldn't divide two inequalities directly.

- 2 months ago

Yes, you did prove that $\sum_{cyc} \frac{x^4}{8x^3+5y^3} \geq \frac{3(xyz)^{\frac13}}{13}$, but unfortunately, that doesn't quite prove the original problem statement. This is because $\sum_{cyc} \frac{x^4}{8x^3+5y^3} \geq \frac{3(xyz)^{\frac13}}{13}$ can definitely mean $\frac{x+y+z}{13} \geq \sum_{cyc} \frac{x^4}{8x^3+5y^3} \geq \frac{3(xyz)^{\frac13}}{13}$.

I'm gonna give a simple example. Suppose that you want to prove $x > y$. You DO know that $y>z$. And somehow you do manage to prove that $x>z$. Does it mean that $x>y$?

No! The conditions $y>z$ and $x>z$ does not imply that $x>y$. See: If $x=2,y=3,z=1$: $y=3>z=1$ TRUE, and $x=2>z=1$ TRUE, however, $x=2!

- 2 weeks, 2 days ago

Yes, I agree with you.. Thank you for correcting me.

- 2 weeks, 2 days ago

There, if you're interested in more inequalities :) https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf

- 2 weeks, 2 days ago

I will surely access it.

- 2 weeks, 2 days ago