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

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 $$a or $$b.

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

## 1 Answer

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.