# Given three positive numbers $a,b,c\in R_{+}^{*}$ prove the following inequality

If $$\ abc=1$$ then prove that $$\sum_{cyc}\frac{1}{3-a+a^{6}}≤1$$ where $$a,b,c>0$$

I think this inequality can be proved by holder ?

My attempt using $$\ am-gm$$

$$3-a+a^{6}≥3-a \quad(etc)$$

$$\sum_{cyc}\frac{1}{3-a+a^{6}}≤\displaystyle\sum_{cyc}\frac{1}{3-a}$$

Now I will get tow case if $$a>3$$ and if $$a<3$$.

If $$a>3,$$ the inequality is true.

Now if $$a<3$$

Using : $$3-a>0$$ (etc)

So: $$\sum_{cyc}\frac{1}{3-a}≤1$$

Is my work correct?

• What happens to your proof when $a=b=c=1$? – Mindlack Jul 17 at 9:01
• @Mindlack Which step ? – Thê Kîng Jul 17 at 9:03
• The bound $\displaystyle\sum_{cyc}\frac{1}{3-a+a^{6}}≤\displaystyle\sum_{cyc}\frac{1}{3-a}$ is not good enough because for $a=b=c=1$ we get $\sum_{cyc}\frac1{3-a} = \frac32 > 1$. – mechanodroid Jul 17 at 9:14

$$\sum_{cyc}\frac{1}{a^6-a+3}\leq1$$ it's $$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}\right)\geq0$$ or

$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}-\frac{5}{9}\ln{a}\right)\geq0.$$ Let $$f(x)=\frac{1}{3}-\frac{1}{x^6-x+3}-\frac{5}{9}\ln{x}.$$ Thus, $$f'(x)=\tfrac{(1-x)(5x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x^4-29x^3-29x^2-24x-45)}{9x(x^6-x+3)^2}.$$ We see that the polynomial $$x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x^4-29x^3-29x^2-24x-45$$

has only one changing of coefficients sign, which by the Descartes's rule https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs

says that this polynomial has unique positive root $$x_1$$ and easy to see that $$x_1>1$$.

For $$x=1$$ our $$f$$ has a local minimum and for $$x=x_1$$ our $$f$$ has a local maximum,

$$f(x_1)>0$$ and $$f$$ decreases on $$[x_1,+\infty).$$

Since $$\lim\limits_{x\rightarrow+\infty}f(x)=-\infty,$$ we obtain that $$f$$ has an unique root $$x_0$$ on $$[x_1,+\infty)$$.

By calculator easy to see that $$x_0=1.696...$$ and since $$f(1)=0$$,

we got that our inequality is proven for $$\max\{a,b,c\}\leq1.696.$$

Now, let $$a>1.696.$$

Thus, by AM-GM $$\sum_{cyc}\frac{1}{a^6-a+3}<\frac{1}{1.696^6-1.696+3}+\frac{2}{3-\frac{5}{6\sqrt[5]6}}=0.867...<1$$ and we are done!