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.
Here is one way, though not very elementary. The inequality is cyclic and on clearing denominators, will be of degree $3$. By the "CD3-improved" theorem on page 315 of Pham Kim Hung's book, the inequality will hold iff
For the 1st condition, $\displaystyle a=b=c \implies \frac3{15a} \ge \frac{\sqrt3}5 \implies a \le \tfrac1{\sqrt3}$ which is true as the constraint ensures $a^2 = \frac13$.
Similarly for the 2nd condition, $c=0 \implies b = 1/a$ and we need to show $$\frac1{3a+5b}+\frac1{3b+7a}+\frac1{5a+7b} \le \frac{\sqrt3}5$$ which can be after some jugglery shown equivalent to $$105 \left(a-\tfrac{71}{42 \sqrt3}\right)^2 a^4+\tfrac{74839}{252} \left(a-\tfrac{35700 \sqrt3}{74839}\right)^2 a^2+\tfrac{20376025}{74839} \left(a-\tfrac{1421941}{1630082 \sqrt3}\right)^2+\tfrac{1036164125}{9780492} \ge 0$$
P.S. Will see if a more satisfying way is possible, or perhaps someone will come up with one.
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 $$ \frac{1}{3a+5b+7c} + \frac{1}{3b+5c+7a} + \frac{1}{3c+5a+7b}\le\frac{3}{15c}=\frac{\sqrt3}{5}. $$
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.