Prove $\sum\limits_{cyc} \frac{1- x_{1}}{5\,{x_{1}}^{4}- 9\,{x_{1}}^{2}- 450}\geq 0$

Question

With $\sum\limits_{i= 1}^{3}x_{i}= 3,\,\,x_{i}\geq 0$, prove that:

$$\sum\limits_{cyc} \frac{1- x_{1}}{5\,{x_{1}}^{4}- 9\,{x_{1}}^{2}- 450}\geq 0$$

Case $x_{i}\geq \frac{2}{\sqrt{5}}$, we have:

$$\sum \left [ \frac{1- x}{5\,x^{4}- 9\,x^{2}- 450}- \frac{1}{454}\left ( x- 1 \right ) \right ]= \sum \frac{\left ( x- 1 \right )^{2}\left ( x+ 1 \right )\left ( 5\,x^{2}- 4 \right )}{- 454\left ( 5\,x^{4}- 9\,x^{2}- 450 \right )}\geq 0$$

Case $0\leq x_{1},\,x_{2}\leq \frac{2}{\sqrt{5}},\,x_{3}\geq 3- \frac{4}{\sqrt{5}}$, I tried to find $k$ such that:

$$\frac{1- x}{5\,x^{4}- 9\,x^{2}- 450}- k\left ( x- 1 \right )\geq 0 \,\,\forall x \in \left [ 0,\,\frac{2}{\sqrt{5}} \right ] \cup \left [ 3- \frac{4}{\sqrt{5}},\,3 \right ]$$

by the way like the post were here: https://diendantoanhoc.net/topic/183270-cmr-sum-frac1a2-geq-sum-a2/#entry711655

But without success! Please help me to find $k$ and solve another cases!

Answer 1

We see that $$450+9x_i^2-5x_i^4\geq450-5\cdot81>0.$$

Let $x_1=a$, $x_2=b$ and $x_3=c$ and $a\geq b\geq c$.

We see that $$450+9b^2-5b^4\geq450+9a^2-5a^4$$ it's $$5(a^4-b^4)-9(a^2-b^2)\geq0$$ or $$5(a^2+b^2)-(a+b+c)^2\geq0$$ or $$4a^2+4b^2-c^2-2ab-2ac-2bc\geq0,$$ which is obviously true.

Thus, we need to prove that $$\sum_{cyc}\frac{a-1}{450+9a^2-5a^4}\geq0$$ or $$\sum_{cyc}\frac{2a-b-c}{450+9a^2-5a^4}\geq0$$ or $$\sum_{cyc}\frac{a-b-(c-a)}{450+9a^2-5a^4}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{1}{450+9a^2-5a^4}-\frac{1}{450+9b^2-5b^4}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)(5(a^2+b^2)-9)(450+9c^2-5c^4)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)(4a^2+4b^2-c^2-2ab-2ac-2bc)(450+9c^2-5c^4)\geq0.$$ We'll prove that $$4a^2+4c^2-b^2-2ab-2ac-2bc\geq0.$$ Indeed, we need to prove that $$4c^2-2(a+b)c+4a^2-2ab-b^2\geq0,$$ for which it's enough to prove that $$(a+b)^2-4(4a^2-2ab-b^2)\leq0$$ or $$(a-b)(3a+b)\geq0,$$ which is obvious.

Id est, $$b^2\sum_{cyc}(a-b)^2(a+b)(4a^2+4b^2-c^2-2ab-2ac-2bc)(450+9c^2-5c^4)\geq$$ $$\geq b^2(a-c)^2(a+c)(4a^2+4c^2-b^2-2ab-2ac-2bc)(450+9b^2-5b^4)+$$ $$+b^2(b-c)^2(b+c)(4b^2+4c^2-a^2-2ab-2ac-2bc)(450+9a^2-5a^4)\geq$$ $$\geq a^2(b-c)^2(a+c)(4a^2+4c^2-b^2-2ab-2ac-2bc)(450+9a^2-5a^4)+$$ $$+b^2(b-c)^2(b+c)(4b^2+4c^2-a^2-2ab-2ac-2bc)(450+9a^2-5a^4)=$$ $$=(b-c)^2(450+9a^2-5a^4)(a^2(a+c)(4a^2+4c^2-b^2-2ab-2ac-2bc)+$$ $$+b^2(b+c)(4b^2+4c^2-a^2-2ab-2ac-2bc))=$$ $$=(b-c)^2(450+9a^2-5a^4)(4(a^2+b^2)c^3+2(a-b)^2(a+b)c^2+$$ $$+2(a^4-2a^3b-a^2b^2-2ab^3+b^4)c+4a^5-2a^4b-a^3b^2-a^2b^3-2ab^4+4b^5)\geq$$ $$\geq(b-c)^2(450+9a^2-5a^4)\left(2(a+b)^2c^3-2\cdot\frac{3}{16}(a+b)^4c+\frac{2}{32}(a+b)^5\right)=$$ $$=\frac{1}{16}(b-c)^2(450+9a^2-5a^4)(a+b)^2(32c^3-6(a+b)^2c+(a+b)^3)=$$

$$=\frac{1}{16}(b-c)^2(450+9a^2-5a^4)(a+b)^2\left(32c^3+\frac{1}{2}(a+b)^3+\frac{1}{2}(a+b)^3-6(a+b)^2c\right)\geq$$ $$\geq\frac{1}{16}(b-c)^2(450+9a^2-5a^4)(a+b)^2\left(3\sqrt[3]{32c^3\cdot\left(\frac{1}{2}(a+b)^3\right)^2}-6(a+b)^2c\right)=0.$$ Done!

Also, the Vasc's RCF Theorem helps here.

By this way finally we obtain $$(3-a)(5a-6)^2(a-1)^2\geq0.$$