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, 10 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.

A Former Brilliant Member - 1 year, 10 months ago

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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 - 1 year, 10 months ago

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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 - 1 year, 10 months ago

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

Hua Zhi Vee - 1 year, 10 months ago

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@Hua Zhi Vee Thank you, for the link

Nikola Alfredi - 7 months, 4 weeks ago

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Hey,

cyclicx48x3+5y3x+y+z133(xyz)1313\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    cyclicx48x3+5y33(xyz)1313\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,z0\ \ x,y,z \geq 0

Nikola Alfredi - 7 months, 4 weeks ago

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SOLUTION

cyclicx48x3+5y3x+y+z133(xyz)1313\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    cyclicx48x3+5y3×13(x3+y3+z3)(x2+y2+z2)2    \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   x2+y2+z2xy+yz+zx\displaystyle \ \ x^2 + y^2 + z^2 \geq xy + yz + zx

Thus,  cyclicx48x3+5y3×(x3+y3+z3)(xy+yz+zx)213\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,   (xy+yz+zx)2139(xyz)4313  \displaystyle \ \ \frac {(xy + yz + zx)^2}{13} \geq \frac {9(xyz)^ \frac {4}{3} }{13} \ \ By AM-GM-HM Inequality

Finally,   (xy+yz+zx)2133(xyz)1313×3xyz  \displaystyle \ \ \frac {(xy + yz + zx)^2}{13} \geq \frac {3(xyz)^ \frac {1}{3} }{13} \times 3xyz \ \

So cyclicx48x3+5y3×(x3+y3+z3)3(xyz)1313×3xyz\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,   x3+y3+z33xyz\displaystyle \ \ x^3 + y^3 +z^3 \geq 3xyz .....................2

Now, as   x,y,z0  \displaystyle \ \ x,y,z \geq 0 \ \ So divide inequality 1 by inequality 2 ,

Hence the final inequality becomes    cyclicx48x3+5y33(xyz)1313\displaystyle \ \ \ \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \geq \frac{3(xyz)^\frac{1}{3} }{13}

If it wasn't metioned that  x,y,z0  \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 - 7 months, 4 weeks ago

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Yes, you did prove that cycx48x3+5y33(xyz)1313 \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 cycx48x3+5y33(xyz)1313 \sum_{cyc} \frac{x^4}{8x^3+5y^3} \geq \frac{3(xyz)^{\frac13}}{13} can definitely mean x+y+z13cycx48x3+5y33(xyz)1313 \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 x > y . You DO know that y>z y>z . And somehow you do manage to prove that x>z x>z . Does it mean that x>yx>y?

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

Hua Zhi Vee - 6 months, 1 week ago

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

Nikola Alfredi - 6 months, 1 week ago

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

Hua Zhi Vee - 6 months, 1 week ago

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I will surely access it.

Nikola Alfredi - 6 months, 1 week ago

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