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 $\ am-gm$

$$3-a+a^{6}≥3-a \quad(etc)$$


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?

  • 2
    $\begingroup$ What happens to your proof when $a=b=c=1$? $\endgroup$ – Mindlack Jul 17 at 9:01
  • $\begingroup$ @Mindlack Which step ? $\endgroup$ – Thê Kîng Jul 17 at 9:03
  • 3
    $\begingroup$ 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 $\sum_{cyc}\frac1{3-a} = \frac32 > 1$. $\endgroup$ – 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!


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