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While trying to prove

If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$

I decided to try to use the technique of letting $a+b+c=3u, ab+bc+ca=3v^2, abc=w^3$ which @Michael Rozenberg used to good effect on If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

The key reasoning is that you obtain some function $g(u,v,w)$ which would need to be negative at any point where the inequality would be false, and consider that as a function of $w$, holding $u$ and $v$ fixed. Then if that $f(w)$ is a decreasing function of $w$, it is enough to prove the original inequality for the maximal possible value of $w$ consistent with our constraints. Since $-3u$ and $3v^2$ are the quadratic and linear coefficients in the polynomial $(x-a)(x-b)(x-c)$, the maximum possible $w$ allowing all of $a,b,c$ to be real comes at a point where two of those are equal.

And once you show that if the inequality is violated it must be violated with two of $a,b,c$ being equal, that can make the proof more tractable.

In the problem at hand (rescaling so that $a+b+c=3abc$ and the number on the RHS is $\frac38$), the expression which if proved non-negative is equivalent to the inequality is

$$ 260w^3 +252uv^2-168u^2-344v^2 -21\sqrt{81u^2v^4-108v^6-108u^3w^3+162uv^2w^3-27w^6} $$ Now for fixed "arbitrary" $u,v$ (subject to the requirement $u\geq v\geq 1$ so that the constraint is feasible) I can't easily show this is an increasing function of $w$.

The trick I want to do is to use the constraint (which is $3w^3=3u$ or $w^3=u$) at one spot to replace $162uv^2w^3$ by $162u^2v^2$. Once this is done, the resulting $f(w)$ is trivially an increasing function of $w$. If I do this, then does what I called the "key reasoning" properly apply, showing the "worst" point for the inequality comes when two of $a,b,c$ are equal?

Or have I "cheated" by replacing $w^3$ in one place but not in others?


EDIT

Corrected typo caught by Macavity.

Question still stands; I'm going to test the validity somewhat by seeing whether my reasoning implies something false when replacing the $7a+b$ by $8a+b$ (where the minimum appears to be attained at a point where all three of $a,b,c$ are different).

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  • $\begingroup$ You can not assume $w^3=u$ because by the given $w^3=3u$. If we want to think about monotony of the function we nee to delete an effect of the condition. For which we need to make our inequality homogeneous. $\endgroup$ – Michael Rozenberg Aug 29 '17 at 1:58
  • $\begingroup$ Assuming you have rescaled correctly, you could put $w^3=u$ of course in one or more terms. Do not see why the steps in the general reasoning would not hold - at the very least it establishes an upper bound on the maximum of $w^3$ for a relaxed constraint situation, which leaves you to show there is equality possible in the end. However not sure if you have done the sqrt term on RHS correctly - do you really get a $v^3$ in the first term under radical sign?? $\endgroup$ – Macavity Aug 30 '17 at 7:08
  • $\begingroup$ @macavity Yes, that was a typo, I am correcting it. $\endgroup$ – Mark Fischler Aug 30 '17 at 15:53
  • $\begingroup$ @ Michael Rozenberg SInce the constraint (in my normalization) is $a+b+c = 3abc$ that is $3u=3w^3$ $\endgroup$ – Mark Fischler Aug 30 '17 at 16:00

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