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The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following inequality."

$$\Sigma_{cyc}\left(\frac{a}{b-c}-3\right)^4\ge193$$

Full expanding gives 12-degree polynomial with about 90 terms. It starts with $\Sigma_{cyc}(a^{12}-16a^{11}b+8a^{11}c)$ and it does not look good for Muirhead or Schur.

Also I tried substitution of $\frac{a}{b-c}=x$, $\frac{b}{c-a}=y$ and $\frac{c}{a-b}=z$. Then by $uvw$, it suffices to show when $x=y$ (See answer to linked question for details). That is, $\frac{a}{b-c}=\frac{b}{c-a}$ or $c=\frac{a^2+b^2}{a+b}$, therefore either $a<c<b$ or $b<c<a$.

Given the constraints, it is clear that $x>0$ and substituting $y=x$, $z=-\frac{1+x^2}{2x}$ gives nonnegative polynomial for $0<x$ (which is not nonnegative polynomial for all $x$).

However, it looks like I cannot deduce $x>0$ from the fact it is enough to consider $x=y$.

How can I prove it? Thank you!

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It remains to make two steps only.

  1. For $$\frac{a}{b-c}=\frac{b}{c-a}$$ or $$c=\frac{a^2+b^2}{a+b}$$ it's enough to prove that $$2\left(\frac{a}{b-\frac{a^2+b^2}{a+b}}-3\right)^2+\left(\frac{a^2+b^2}{a^2-b^2}-3\right)^4\geq193.$$ Now, let $a=tb$.

Thus, we need to prove that $$2\left(\frac{t}{b-\frac{t^2+1}{t+1}}-3\right)^2+\left(\frac{t^2+1}{t^2-1}-3\right)^4\geq193$$ or $$335t^8+1024t^7+338t^6-1280t^5-742t^4+640t^3+196t^2-128t+95\geq0,$$ which is obviously true for $t>0$.

  1. For $w^3\rightarrow0$ let $\frac{c}{a-b}\rightarrow0$.

Thus, we need to prove that $$\left(\frac{a}{b}-3\right)^4+\left(\frac{b}{-a}-3\right)^4+81\geq193$$ or $$(a^2-4ab-b^2)^2(a^4-4a^3b+8a^2b^2+4ab^3+b^4)\geq0,$$ which is obvious again.

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