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.

Fact 1: If $x, y, z > 0$ with $-71(x^2 + y^2 + z^2) + 83(xy + yz + zx) = 2700$, then
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \le \frac{\sqrt 3}{5}.$$
(The proof is given at the end.)
Now, let
\begin{align*}
 x &= 3a + 5b + 7c, \\
 y &= 3b + 5c + 7a, \\
 z &= 3c + 5a + 7b.
\end{align*}
We have $x, y, z > 0$ and
$$-71(x^2 + y^2 + z^2) + 83(xy + yz + zx)
= 2700(ab + bc + ca) = 2700.$$
By Fact 1, we have
$$\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}
\le\frac{\sqrt{3}}{5}.$$
We are done.
$\phantom{2}$

Proof of Fact 1:
Letting $x = 5u\sqrt 3, y = 5v\sqrt 3, z = 5w \sqrt 3$, it suffices to prove the following:
Fact 2: If $u, v, w > 0$ with $-71(u^2 + v^2 + w^2) + 83(uv + vw + wu) = 36$, then
$$3uvw \ge uv + vw + wu.$$
Let $p = u + v + w, q = uv + vw + wu, r = uvw$.
The condition $-71(u^2 + v^2 + w^2) + 83(uv + vw + wu) = 36$
is written as $-71(p^2 - 2q) + 83q = 36$ or
$$q = \frac{36 + 71p^2}{225}. \tag{1}$$
Using $p^2 \ge 3q$ and (1), we have
$$p \ge 3.$$
It suffices to prove that
$$3r \ge q.$$
Using degree three Schur, we have
$$r \ge \frac{4pq - p^3}{9}.$$
It suffices to prove that
$$3 \cdot \frac{4pq - p^3}{9} \ge q$$
or (using (1))
$$\frac{1}{225}(p - 3)(59p^2 - 36p + 36) \ge 0$$
which is true.
We are done.
