<s>[INST] Is the maximum of $\sum\limits_{\mathrm{cyc}} \frac{1}{3a + 5b + 7c}$ equal to $\frac{\sqrt{3}}{5}$? 
Let $a,b,c>0$ such $ab+bc+ac=1$. Show that
$$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{5}.$$
(Note: $\mathrm{LHS} = \mathrm{RHS}$ when $a = b = c = 1/\sqrt 3$.)

since dear Mac sir,he solve with inequality $\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}\le\frac{\sqrt{3}}{4}$
use same methods,I think
$$(3a+5b+7c)^2\ge 75=75(ab+bc+ac)?$$
I would appreciate any help. Thanks in advance.
 [/INST] You can use Lagrange multipliers. Let's denote
$$L(a, b, c) = \frac{1}{3a+5b+7c} + \frac{1}{3b+5c+7a} + \frac{1}{3c+5a+7b}-\lambda(ab+ac+bc-1);$$
we have system
$$\left\{\frac{\partial L}{\partial a}=0, \frac{\partial L}{\partial b}=0, \frac{\partial L}{\partial c}=0\right\},$$
or
$$
\left\{
\begin{aligned}
\frac{3}{(3a+5b+7c)^2} + \frac{7}{(3b+5c+7a)^2} + \frac{5}{(3c+5a+7b)^2}+\lambda(b+c)=0\\
\frac{5}{(3a+5b+7c)^2} + \frac{3}{(3b+5c+7a)^2} + \frac{7}{(3c+5a+7b)^2}+\lambda(a+c)=0\\
\frac{7}{(3a+5b+7c)^2} + \frac{5}{(3b+5c+7a)^2} + \frac{3}{(3c+5a+7b)^2}+\lambda(b+c)=0
\end{aligned}
\right.
$$
It's linear system for squares in denominators; solution is
$$
\left\{
\begin{aligned}
\frac{1}{(3a+5b+7c)^2}=\frac{90}{\lambda}(11c-4b-19a)\\
\frac{1}{(3b+5c+7a)^2}=\frac{90}{\lambda}(11a-4c-19b)\\
\frac{1}{(3c+5a+7b)^2}=\frac{90}{\lambda}(11b-4a-19c)
\end{aligned}
\right.,
$$
or if you prefer
$$
\left\{
\begin{aligned}
(3a+5b+7c)^2(11c-4b-19a)=\lambda/90\\
(3b+5c+7a)^2(11a-4c-19b)=\lambda/90\\
(3c+5a+7b)^2(11b-4a-19c)=\lambda/90
\end{aligned}
\right..
$$
We can solve this system in Maple, or to find solution in a form $a=b=c$:
$$
3c^2=1\Longrightarrow a=b=c=1/\sqrt3,
$$
and $\lambda = -243000c^3$ from the last system (other solutions are complex). Substituting it into your function, we have
$$
\frac{1}{3a+5b+7c} + \frac{1}{3b+5c+7a} + \frac{1}{3c+5a+7b}\le\frac{3}{15c}=\frac{\sqrt3}{5}.
$$
</s>