Inequality Confusion, please guide me.

Asked Mar 09 '20 at 18:21

Active Mar 09 '20 at 18:21

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Question:

We need to prove the following inequality :-

$$\displaystyle \sum_{cyclic} \frac{x^4}{8x^3 + 5y^3} \geq \frac{x+y+z}{13} \ \ \ \ \ \ \ \ \ \ \ x,y,z \in \mathbb{R}^+ $$

As we know by AM-GM that,

$$\displaystyle \frac{x+y+z}{13} \geq \frac{3(xyz)^\frac{1}{3} }{13} $$ Can we say it is similar to prove this

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

instead of the original equation.

Because then,

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)^2 \geq 9(xyz)^\frac{4}{3} $

This will result in

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

Now we can tell $\displaystyle {(x^3 + y^3 + z^3)} \geq {3(xyz)} \ \ \ $ (always true) i.e it is equivalent to say that $\displaystyle \sum_{cyclic} (\frac{x^4}{8x^3 + 5y^3}) \geq \frac{3(xyz)^\frac{1}{3} }{13} .$

Am I correct ?

asked Mar 09 '20 at 18:21

Nikola Alfredi

If you find a lower bound to the RHS and prove that LHS is greater than that bound, it does not mean that LHS is greater than RHS. – LHF Mar 09 '20 at 18:31
Ok, Some other way to do this ? – Nikola Alfredi Mar 09 '20 at 18:32
And why is it that ? – Nikola Alfredi Mar 09 '20 at 18:32
the thing you mentioned as similar to prove is not correct. AM>=GM but if you prove x^4/(8x^3 + 5y^3) >= GM then it might even be that AM >= x^4/(8x^3 + 5y^3). AM can be even more. – user693540 Mar 09 '20 at 18:33
Because if 4>3 and 5>3, it doesn't mean 4>5. – LHF Mar 09 '20 at 18:34
@user693540 Please clearify the reason ... – Nikola Alfredi Mar 09 '20 at 18:34
@Atticus Ok, I got your point. That was my confusion. – Nikola Alfredi Mar 09 '20 at 18:37
Can you please suggest some other way ? – Nikola Alfredi Mar 09 '20 at 18:38
If you're just learning inequalities, I suggest to put this inequality aside for some time. I think it's very hard. – LHF Mar 09 '20 at 18:39
No, I am quite handy with them... – Nikola Alfredi Mar 09 '20 at 18:40
The only thing is I am not able to relate with the problem. I am confused which inequality to use. – Nikola Alfredi Mar 09 '20 at 18:41
I am aware about Jesen's Inequality, Holder's inequality, Muirhead's Inequality,Chebyshev's Inequality.. – Nikola Alfredi Mar 09 '20 at 18:47
1

@NikolaAlfredi, The problem is I highly doubt any classical inequality works for this question. It would probably need some brute-force way with computer software. What is the source? – LHF Mar 09 '20 at 18:49
Ok, do you know something about it ? – Nikola Alfredi Mar 09 '20 at 18:50
[Source]
: https://brilliant.org/discussions/thread/olympiad-inequality/ – Nikola Alfredi Mar 09 '20 at 18:55
2

There's a refference in your link to the following question: https://math.stackexchange.com/questions/1775572/olympiad-inequality-sum-limits-cyc-fracx48x35y3-geqslant-fracxy. That pretty much clears it up what a hard question this is. – LHF Mar 09 '20 at 18:58
Thank you for the help, I really appreciate it. – Nikola Alfredi Mar 09 '20 at 19:05
@NikolaAlfredi If you know that $A\geq C$ and $B\geq C$ then you still don't know if $A\geq B$ is true so this is why your approach doesn't work – Maximilian Janisch Mar 09 '20 at 22:26
Thank you for guiding. – Nikola Alfredi Mar 10 '20 at 09:16

Inequality Confusion, please guide me.

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