Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$

Question

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$

I tried homogenization and the BW (https://artofproblemsolving.com/community/c6h522084),

but it does not work.

Indeed, let $a=\frac{x}{y}$, $b=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.

Hence, $c=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}\leq\frac{3}{2},$$ which has a problem around $(x,y,z)=(7,5,6)$.

For these values $$\frac{3}{2}-\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}=0.0075...$$ I tried also TL, uvw, C-S, Lagrange multipliers and more, but without success.

Also, Vasc's Theorems don't help.

Also, the following method does not help here. Find the maximum of the expression

Because the inequality $\frac{x}{x^{11}+1}\leq\frac{3(a^9+1)}{4(a^{18}+a^9+1)}$ is wrong.

Answer 1

Define $$ f(a,\lambda) = -\frac{a}{a^{11}+1} + \lambda \log(a) + \frac{1}{2} $$ Then, for any choice of $\lambda$, $$ f(a,\lambda) + f(b,\lambda) + f(c,\lambda) = -\frac{a}{a^{11}+1} -\frac{b}{b^{11}+1} -\frac{c}{c^{11}+1} + \frac{3}{2} $$ and we need to show that this is $\ge 0$.

It suffices to show that, for some $\lambda^*$ and for all $a$, $f(a, \lambda^*) \ge 0$.

Clearly, for any lambda, $f(a=1,\lambda) = 0$. In order to keep $f(a,\lambda) $ positive for $a >1$ and $a <1$, we demand

$$ 0 = \frac{d f(a,\lambda)}{d a}|_{a=1} $$

which results in $\lambda^* = - \frac94$. We therefore investigate

$$ f(a,\lambda^*) = -\frac{a}{a^{11}+1} -\frac{9}{4} \log(a) + \frac{1}{2} $$ By inspection, we have that $f(a,\lambda^*) \ge 0$ for $a\in (0, 1.1]$. So the inequality is obeyed at least for $a,b,c < 1.1$, and it remains to be shown that the inequality is obeyed outside this specification.

This gives rise to three cases:

case 1: $a,b,c > 1.1$. This is not possible since $abc = 1$.

case 2: $a < 1.1$ ; $b,c > 1.1$. Now observe two facts:

1. By inspection, $ \frac{a}{a^{11}+1} < 0.75$ for any $a$.

2. For $b > 1.1$, $ \frac{b}{b^{11}+1} \le \frac{1.1}{1.1^{11}+1} \simeq 0.2855$ since $ \frac{b}{b^{11}+1}$ is falling for $b > 1.1$.

Hence, in case 2, $ \frac{a}{a^{11}+1} + \frac{b}{b^{11}+1}+ \frac{c}{c^{11}+1} < 0.75 + 2\cdot 0.2855 = 1.3210 < \frac32$ which proves case 2.

case 3: $a,b < 1.1$ ; $c > 1.1$. Here $abc = 1$ requires $a\cdot b =1/c < 1.1^{-1} = 0.909$. Also note that, for some given $c$, $1/(1.1 c) <a<1.1$ in order to observe $a,b < 1.1$. Following case 2, we have that $f(c) = \frac{c}{c^{11}+1} $ is falling with $c$. These conditions could be further exploited (this has not yet been pursued in the comments).

As Martin R. poined out, the maximum will be attained at a point where at least two out of $a,b,c$ equal. In this case, this would be $a=b$. So we can consider proving $$ g(a) = \frac32 - \frac{2 a}{a^{11}+1} - \frac{a^{-2}}{a^{-22}+1} \ge 0 $$ for $a < 1/\sqrt{1.1} \simeq 0.9535$.

Note that in this range, the minimum of $g(a)$ occurs at $a^*\simeq 0.8385$ and has a value of $g(a^*) \simeq 0.00525$. Other than this inspection of the function $g(a)$, I couldn't offer a better proof.

Answer 2

In principle it is, if not always, almost very often that such a problem can be solved using techniques from optimization. For instance one can consider the following maximization problem: \begin{equation} \max_{a,b,c\in C} f(a,b,c) \end{equation} where the constraint set $C:=\{a,b,c\in\mathbb{R}_+:abc=1\}$ and $f(a,b,c):=\sum_{cyc}a/(a^{11}+1)$. If one shows that $3/2$ is the maximum value $f(a,b,c)$ attains in $C$ then this solves the inequality problem. We will follow the same idea however first we transform the given inequality into an equivalent form together with an appropriate constraint which makes it easier to solve it as a maximization problem. The original inequality is given by: \begin{equation} \frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leqslant \frac{3}{2} \end{equation} and $abc=1$. After proper rearrangements this last inequality is equivalent to: $$2(a(b^{11}+1)(c^{11}+1)+b(a^{11}+1)(c^{11}+1)+c(a^{11}+1)(b^{11}+1))\leqslant 3(a^{11}+1)(b^{11}+1)(c^{11}+1)$$ or equivalently: $$2\Big((b^{10}+\frac{1}{b})(c^{10}+\frac{1}{c})+(a^{10}+\frac{1}{a})(c^{10}+\frac{1}{c})+(a^{10}+\frac{1}{a})(b^{10}+\frac{1}{b})\Big)\leqslant 3(a^{10}+\frac{1}{a})(b^{10}+\frac{1}{b})(c^{10}+\frac{1}{c})$$ Let $f(x):=x^{10}+1/x$ then the last inequality is the same as: $$\frac{1}{f(a)}+\frac{1}{f(b)}+\frac{1}{f(c)}\leqslant\frac{3}{2}$$ It is sufficient to look at the problem: $$\max_{a,b,c}F(a,b,c):=\frac{1}{f(a)}+\frac{1}{f(b)}+\frac{1}{f(c)}$$ subject to $abc=1$. The Lagrangian for this problem is: $$L(a,b,c,\lambda):=F(a,b,c)-\lambda(1-abc)$$ From the first order conditions we get the following equations: $$\frac{f'(a)}{f^2(a)}=\lambda bc\Leftrightarrow a\frac{f'(a)}{f^2(a)}=\lambda \\ \frac{f'(b)}{f^2(b)}=\lambda ac\Leftrightarrow b\frac{f'(b)}{f^2(b)}=\lambda \\ \frac{f'(c)}{f^2(c)}=\lambda ab\Leftrightarrow c\frac{f'(c)}{f^2(c)}=\lambda $$ A possible obvious solution to this system is $a=b=c=1$ and $\lambda=9/4$. If one calculates the Hessian of $L(a,b,c,\lambda)$ (bordered Hessian) we get: $$\text{Hess}_L(a,b,c,\lambda)= \begin{bmatrix} 0 & bc & ac & ab\\ bc & F_{aa} & \lambda c& \lambda b \\ ac &\lambda c & F_{bb} & \lambda a\\ ab &\lambda b & \lambda a & F_{cc} \end{bmatrix} $$ where $F_{xx}:=-\Big[(f''(x)f^2(x)-2(f'(x))^2f(x))/f^4(x)\Big]$. Evaluating at $(1,1,1,9/4)$ gives: $$\text{Hess}_L(1,1,1,9/4)= \begin{bmatrix} 0 & 1 & 1 & 1\\ 1 & -11/4 & 9/4& 9/4 \\ 1 &9/4 & -11/4 & 9/4\\ 1 &9/4 & 9/4 & -11/4 \end{bmatrix} $$ From this follows $\det \text{Hess}_L(1,1,1,9/4)=-75$ and $\det M_L(1,1,1,9/4)=10$ where $M$ is the submatrix $$M_L:=\begin{bmatrix} 0 & 1 & 1 \\ 1 & -11/4 & 9/4 \\ 1 &9/4 & -11/4 \end{bmatrix} $$ These satisfy the second order conditions for our bordered Hessian (the alternating sign condition) for a local maximum at $(1,1,1)$.

Answer 3

We start with the case $a\leq 1$ , $b\leq 1$ , $c\geq 1$ so we have to prove this : $$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq1.5$$ Or with $a\geq 1$, $b\geq 1$ , $c\leq 1$ : $$\frac{a^{10}}{a^{11}+1}+\frac{b^{10}}{b^{11}+1}+\frac{c^{10}}{c^{11}+1}\leq1.5$$ We have the following lemma :

Let $a,b$ be real positive numbers with $a\geq 1$, $b\geq 1$ then we have : $$\frac{a^{10}}{a^{11}+1}+\frac{b^{10}}{b^{11}+1}\leq \frac{a+b}{ab}\frac{(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}$$

Proof :

It's just the inequality of Jensen apply to $f(x)$ wich is concave for $x\geq 1$ :

$f(x)=\frac{x^{11}}{x^{11}+1}$

With coefficient :

$\alpha_1=\frac{1}{a}\frac{ab}{a+b}$

And

$\alpha_2=\frac{1}{b}\frac{ab}{a+b}$

So we have to prove :

$$\frac{c^{10}}{c^{11}+1}+\frac{a+b}{ab}\frac{(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\leq 1.5$$

We have this other lemma :

$$\frac{c^{10}}{c^{11}+1}=\frac{ab}{(ab)^{11}+1}\leq \frac{(\frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1} $$

Proof :

It's easy to show this because $f(x)=\frac{x}{x^{11}+1}$ is decreasing for $x\geq 1$

It's remains to prove : $$(\frac{2ab}{a+b})^{2}\leq ab $$ Or : $$ab\leq 0.5(a+b)^2 $$

Wich is obvious.

So we have to prove this :

$$\frac{(\frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1}+\frac{a+b}{ab}\frac{(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\leq 1.5$$

But we put :

$x=\frac{2ab}{a+b}$

We get :

$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 1.5$$

Wich is true !

The other case wich is $a\leq 1$ , $b\geq 1$ , $c\geq 1$ and :

$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq1.5$$

Is easy to show because :

$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{a}{a^{11}+1}+\frac{b^{10}}{b^{11}+1}+\frac{c}{c^{11}+1}$$

Done !

Edit :

We prove that for $x\geq 1$:

$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 1.5$$

First we prove that :

$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 2$$

We have the following identity :

$$\frac{2}{x}-\frac{2}{x(x^{11}+1)}+\frac{1}{x^{20}}-\frac{1}{x^{20}(x^{22}+1)}=\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}$$

So we have to prove this :

$$\frac{2}{x}+\frac{1}{x^{20}}\leq\frac{2}{x(x^{11}+1)}+\frac{1}{x^{20}(x^{22}+1)}+2$$

Wich is obvious because we have :

$$\frac{2}{x}\leq 2$$ And $$\frac{1}{x^{20}}\leq \frac{2}{x(x^{11}+1)}$$ So we have proved : $$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 2$$ Or $$\frac{x^{10}}{x^{11}+1}+\frac{0.5x^2}{x^{22}+1}\leq 1$$

Now we put :

$f(x)=\frac{x^{10}}{x^{11}+1}+\frac{0.5x^2}{x^{22}+1}$

$g(x)=\frac{x^{10}}{x^{11}+1}$

$h(x)=\frac{0.5x^2}{x^{22}+1}$

And we want to prove this :

$$\frac{f(x)^{11}}{f(x)^{11}+1}\leq \frac{0.75^{11}}{0.75^{11}+1}$$

First of all the function $u(x)=\frac{x^{11}}{x^{11}+1}$ is convex for $x\leq 1$ so we put $x=\frac{1}{y}$ and :

$g(x)=\frac{x^{10}}{x^{11}+1}=\frac{y}{y^{11}+1}=l(y)$

$h(x)=\frac{0.5x^2}{x^{22}+1}=\frac{0.5y^{20}}{y^{22}+1}=p(y)$

$$f(x)=q(y)=\frac{y}{y^{11}+1}+\frac{0.5y^{20}}{y^{22}+1}$$

It remains to prove :

$$\frac{q(y)^{11}}{q(y)^{11}+1}\leq \frac{0.75^{11}}{0.75^{11}+1}$$

We can apply Jensen's inequality to $u(x)$ (because all the value are inferior to one) to get :

$$\frac{q(y)^{11}}{q(y)^{11}+1}\leq [\frac{\alpha}{\alpha+\beta}\frac{(\frac{l(y)}{\alpha})^{11}}{(\frac{l(y)}{\alpha})^{11}+1}+\frac{\beta}{\alpha+\beta}\frac{(\frac{p(y)}{\beta})^{11}}{(\frac{p(y)}{\beta})^{11}+1}]$$

We put finally :

$$\alpha=\frac{4l(y)}{3}$$ And $$\beta=\frac{4p(y)}{3}$$

To get : $$\frac{q(y)^{11}}{q(y)^{11}+1}\leq \frac{0.75^{11}}{0.75^{11}+1}$$

But the function $u(x)$ is increasing so we have :

$$q(y)\leq \frac{3}{4}$$

Or

$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 1.5$$

Done !

Answer 4

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FUNCTIONS

At first, let us consider some functions of the non-negative argument.

The continuous function $$f(t)=\dfrac{t}{t^{11} + 1}\tag{1a}$$ has the next properties: $$f(0) = f(\infty) = 0,\quad f(x) > 0 \quad \text{if}\quad t\in(0, \infty).\tag{1b}$$ The derivative $$\dfrac{df}{dt} = \dfrac{1 - 10 t^{11}}{(t^{11} + 1)^2}\tag{1c}$$ has the single root $$t_m = \dfrac1{\sqrt[11]{10}} \approx 0.81113,\quad f'(t_m)= 0,\quad f(t_m) = f_m = \dfrac1{11}10^{10/11}\approx 0.73739.\tag{1d}$$ These mean that the function $f(t)$ is positive for all positive $t,$ with the maximum $f_m$ in the point $t_m$ and the values range $f(x) \in [0, f_m].$
Besides, the function $f(t)$ increases monotonically in $[0, t_m)$ and decreases monotonically in $(t_m, \infty).$

The continuous function $$g(u) = 2f\left(u^{-1/11}\right) + f(u^{2/11})\tag{2a},$$ or $$g(u) = \dfrac{2u^{10/11}}{u+1}+\dfrac{u^{2/11}}{u^2+1},\tag{2b}$$ is the positive in the interval $u\in\left[\dfrac1{10}, 10\right].$ The derivative $$g'(u) = \dfrac2{11}\dfrac{10-u}{(u+1)^2}u^{-1/11}-\dfrac2{11}\dfrac{1-10u^2}{u(u^2 + 1)^2}u^{2/11}\tag{2c}.$$ has the roots $$u_0 = 1,\quad u_1 \approx 2.4,\quad u_2\approx 6.933583,$$ wherein $$g(1) = \dfrac32,\quad g(u_1)\approx 1.477,\quad g(u_2) \approx1.49475 < \dfrac32. $$ Therefore, $$g(u) \le \dfrac32\quad \text{ if } u\in\left[\dfrac1{10}, 10\right].\tag{2d}$$

The continuous function $$h(t) = tf'(t) = t\dfrac{1-10t^{11}}{(t^{11}+1)^2}\tag{3a}$$ has the next properties: $$\begin{cases} h(0) = h(t_m) = 0\\ h(t) < 0 \text{ if } t\in(0,t_m)\\ h(t) > 0 \text{ if } t\in(t_m, \infty). \end{cases}\tag{3b}$$ The derivative $$h'(t) = \dfrac{100t^{22} - 141t^{11}+ 1}{(t^{11} + 1)^3}\tag{3c}$$ has the roots $$t_1 = \sqrt[11]{\frac{141-11\sqrt{161}}{200}} = \left(\dfrac{\sqrt{161} - 11}{20}\right)^{2/11} \approx 0.63799 \in (0, t_m),\tag{3d}$$ $$t_2 = \sqrt[11]{\frac{141+11\sqrt{161}}{200}} = \left(\dfrac{\sqrt{161} + 11}{20}\right)^{2/11}\approx 1.0313 \in (t_m, \infty)\tag{3e}.$$ These mean that the function $h(t)$ is positive in $(0, t_m)$ and negative in $(t_m, \infty).$
Besides, it increases monotonically in $(0, t_1)$ and $(t_2, \infty)$ and decreases monotonically in $(t_1, t_2).$

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THE TASK ANALYSIS

The issue inequality can be written in the form of $$\Phi(x, y, z) \le \dfrac32,\tag{4a}$$ where $$\Phi(x, y, z) = f(x) + f(y) + f(z)\tag{4b}.$$ The issue conditions are $$xyz = 1,\quad (x, y, z)\in (0,\infty)^3.\tag{4c}$$

Note that $$f(1) = \dfrac12, \quad f(t) < \dfrac12 \text{ if } t\in(1,\infty).$$ This means that the inequality $(4a)$ becomes exact equality when $x = y = z = 1.$
Also that means that at least one of the values $x,\ y,\ z$ belongs to the interval $(0, 1].$
On the other hand, the conditions $(4c)$ provide at least one of the values $x,\ y,\ z$ belongs to the interval $[1, \infty).$

Let WLOG $0 < x \le y \le z,$ then it suffices to consider cases $y\le 1$ and $y > 1.$

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CASE $\mathbf{0 < x \le y \le 1 \le z.}$

Let us find the maximum of $$F(x, y) = f(x) + f(y) + f\left(\dfrac1{xy}\right),\quad (x, y) \in (0,1]^2.\tag{5a}$$ The nesessary extremum conditions of $F(x, y)$ are $F'_x = F'_y = 0,$ or \begin{cases} f'(x) - \dfrac1{x^2y} f\left(\dfrac1{xy}\right) = 0\\ f'(y) - \dfrac1{xy^2} f\left(\dfrac1{xy}\right) = 0. \end{cases} Taking in account $(3a),$ this system can be presented in the form of $$h(x) = h(y) = h\left(\dfrac1{xy}\right).\tag{5b}$$ In the same time, $\dfrac1{xy} > 1,\ h\left(\dfrac1{xy}\right) < 0,$ so the system $(5b)$ contents the negative values.
Thus, $$t_m < x \le y \le 1 \le z,\quad h(x) = h(y) = h\left(\dfrac1{xy}\right) < 0.\tag{5c}$$ In accordance with $(3d),$ the function $h(x)$ is monotonic in the $[t_m, 1].$ Therefore, the system $(5c)$ leads to $$x = y,\quad z = \dfrac1{x^2},\tag{5d}$$ $$F(x, y) = 2f(x) + f\left(\dfrac1{x^2}\right),$$ and, taking in account $(2a)$, $$F(x,y) = g(x^{-11}),\quad x^{-11}\in[1, 10).$$ In accordance with $(2d),\quad F(x, y,z)\le\dfrac32.$

Thus, the issue inequality is proved in the case $\mathbf{y \le 1.}$

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THE CASE $\mathbf{0 < x \le 1 < y \le z.}$

Let us find the maximum of $$G(y, z) = f\left(\dfrac1{yz}\right) + f(y) + f(z)\quad (y, z) \in (1,\infty)^2.\tag{6b}$$ The nesessary extremum conditions of $G(y, z)$ are $G'_y = F'_z = 0,$ or \begin{cases} - \dfrac1{y^2z} f'\left(\dfrac1{yz}\right) + f'(y) = 0\\ - \dfrac1{yz^2} f'\left(\dfrac1{yz}\right) + f'(z)= 0. \end{cases} Taking in account $(3a)$, this system can be presented in the form of $$h\left(\dfrac1{yz}\right) = h(y) = h(z).\tag{6b}$$ In the same time, $$1 < y \le z\quad\Rightarrow\quad h(y) = h(z) < 0.$$ Thus, $$t_m < \dfrac1{yz} < 1 < y \le z,\quad h\left(\dfrac1{yz}\right) = h(y) = h(z) < 0.\tag{6c}$$ The function $h(t)$ monotonically decreases in $(t_m, t_2)$ and monotonically increases in $(t_2, \infty).$

If $y\le t_2,$ then, taking in account $(4c),$ the values $h(1/yz)$ and $h(y)$ belongs to the decreasing branch of h(t). That leads to contradiction $\dfrac1{yz} = y$ with $1 < y < z,$ and then the system $(6b)$ has not solutions.

Therefore $$t_m < \dfrac1{yz} \le 1 \le r_2 \le y \le z,\quad h\left(\dfrac1{yz}\right) = h(y) = h(z) < 0,$$

These mean that both of the values $h(y) = h(z)$ belongs to the same increasing branch, so $$y = z,\quad x = \dfrac1{z^2},\tag{6d}$$ $$F(x, y, z) = 2f(z) + f\left(\dfrac1{z^2}\right),$$ and, taking in account $(2)$, $$F(x,y,z) = g(z^{-11}),\quad z^{-11}\in\left[\dfrac1{10}, 1\right).$$ In accordance with $(2d),\quad F(x, y,z)\le\dfrac32.$

The issue inequality is proved in the case $\mathbf{y > 1.}$

Thus,
if $xyz =1,\ (x,y,z) \in(0,\infty)^3,$
then $$\boxed{\dfrac x{x^{11}+1} + \dfrac y{y^{11}+1} + \dfrac z{z^{11}+1} \le \dfrac32.}$$

Answer 5

This is probably wrong, but it might provide some ideas.

First observe that $a^nb^nc^n\leq3\;\;\forall n$, which is trivial by AM-GM.

Then, when $x\geq y$, then $\frac1x\leq\frac1y$.

First, expand to get $$a(b^{11}+1)(c^{11}+1)+b(a^{11}+1)(c^{11}+1)+c(a^{11}+1)(b^{11}+1)\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$

so

$$a+b+c+ab(a^{10}+b^{10})+bc(b^{10}+c^{10})+ac(a^{10}+c^{10})+abc(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$

By the first observation $a+b+c\geq3$. Also, by AM-GM, $a^{10}+b^{10}\geq\frac2{c^5}$. And since $ab=\frac1c$, we get

$$3+2\left(\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\right)+(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$

Then, by GM-HM, $$1\geq{3\over{(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over a^{10}b^{10}c^{10}}}$$ so $$(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\geq3$$

Thus we get $$6+2\left(\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\right)\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$

And similarly to above, with GM-HM we get $$\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\geq3$$

Thus we get $$12\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$

And expanding the denominator we get $$12\over2+(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})+(a^{11}+b^{11}+c^{11})$$

And now I'm not sure what to do. Hope this helps.