Show that if $\frac1a+\frac1b+\frac1c = a+b+c$, then $\frac1{3+a}+\frac1{3+c}+\frac1{3+c} \leq\frac34$ 
Show that if  $a,b,c$ are positive reals, and 
  $\frac1a+\frac1b+\frac1c = a+b+c$, then
  $$\frac1{3+a}+\frac1{3+b}+\frac1{3+c} \leq\frac34$$

The corresponding problem replacing the $3$s with $2$ is shown here:
How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$?
The proof is not staggeringly difficult: The idea is to show (the tough part) that if
$\frac1{2+a}+\frac1{2+c}+\frac1{2+c}=1$ then 
$\frac1a+\frac1b+\frac1c \ge a+b+c$. The result then easily follows.
A pair of observations, both under the constraint of $\frac1a+\frac1b+\frac1c = a+b+c$:


*

*If $k\geq 2$ then $\frac1{k+a}+\frac1{k+c}+\frac1{k+c}\leq\frac3{k+1}$. 
This is proven for $k=2$ but I don't see how for $k>2$

*If $0<k<2$ then there are positive  $(a,b,c)$ satisfying the constraint such that $\frac1{k+a}+\frac1{k+c}+\frac1{k+c}>\frac3{k+1}$.
So one would hope that the $k=3$ case, lying farther within the "valid" region, might be easier than $k=2$ but I have not been able to prove it. 
Note that it is not always true, under our constraint, that 
$\frac1{2+a}+\frac1{2+c}+\frac1{2+c}\geq \frac1{3+a}+\frac1{3+c}+\frac1{3+c}+\frac14$, which if true would prove the $k=3$ case immediately.

By C-S
$$\sum_{cyc}\frac{1}{3+a}\leq\frac{1}{(3+1)^2}\sum_{cyc}\left(\frac{3^2}{2+a}+\frac{1^2}{1}\right)=$$
$$=\frac{9}{16}\sum_{cyc}\frac{1}{2+a}+\frac{3}{16}\leq\frac{9}{16}+\frac{3}{16}=\frac{3}{4}$$
and we are done!
