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

#Algebra

Easy Math Editor

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Comments

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

A Former Brilliant Member - 2 years, 7 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.

Hua Zhi Vee - 2 years, 7 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.

A Former Brilliant Member - 2 years, 7 months ago

@A Former Brilliant Member – Ok. I was wondering if it could be solved using inequalities/theorems like these.

Hua Zhi Vee - 2 years, 7 months ago

@Hua Zhi Vee – Thank you, for the link

Nikola Alfredi - 1 year, 4 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 $\ \ x,y,z \geq 0$

Nikola Alfredi - 1 year, 4 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.

Please check if it is correct.... I am in grade 10th

Nikola Alfredi - 1 year, 4 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, $x=2<y=3$ !

Hua Zhi Vee - 1 year, 3 months ago

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

Nikola Alfredi - 1 year, 3 months ago

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

Hua Zhi Vee - 1 year, 3 months ago

I will surely access it.

Nikola Alfredi - 1 year, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$