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.
 A: For the case $k≥3$, $(k\in\mathbb Z)$
Given equality implies $ab+bc+ca=abc(a+b+c)$
We know that $(x+y+z)^2≥3(xy+yz+zx)$
put $x=ab,y=bc,z=ca\Rightarrow (ab+bc+ca)^2\geq3abc(a+b+c)=3(ab+bc+ca)$
$\Rightarrow abc(a+b+c)=ab+bc+ca\geq3$ and $a+b+c\geq3$ (obvious)
$\dfrac {1}{k+a}+\dfrac {1}{k+b}+\dfrac {1}{k+c}≤\dfrac {3}{k+1} \Leftrightarrow\dfrac {3k^2+2k(a+b+c)+(ab+bc+ca)}{k^3+k^2(a+b+c)+k(ab+bc+ca)+abc}\leq\dfrac {3}{k+1}$
$\Leftrightarrow3k^3+3k^2+(2k^2+2k)(a+b+c)+(k+1)(ab+bc+ca)\leq3k^3+3k^2(a+b+c)+3k(ab+bc+ca)+3abc$
$\Leftrightarrow3k^2\leq(k^2-2k)(a+b+c)+(2k-1)(ab+bc+ca)+3abc$
Last inequality is true by Arithmetic-Geometric mean (and by $k^2-2k\geq1$)
$ \underbrace {(a+b+c)+...+(a+b+c)}_{k^2-2k\geq3 \text{ times}}+\underbrace {(ab+bc+ca)+...+(ab+bc+ca)}_{2k-1 \text{ times}}+\underbrace {3abc}_{1 \text{ times}}\geq$
$k^2\sqrt[k^2]{3abc(a+b+c)^{k^2-2k}(ab+bc+ca)^{2k-1}}=k^2\sqrt[k^2]{3(a+b+c)^{k^2-2k-1}(ab+bc+ca)^{2k}}\geq3k^2$
A: 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!
A: This is a follow-on to Michael Rozenberg's answer, generalizing to all real $k>2$, given only the theorem that under the given constraint,
$$
\sum_{\mbox{cyc}}\frac1{2+a}\leq 1
$$
Use (C-S) 
$$ \sum \left(\frac{\alpha_i^2}{x_i}\right) \geq \frac{\left(\sum \alpha_i\right)^2}{\sum x_i}
$$
for the $n=2$ case with
$$
\matrix{ \alpha_1 = 3 & \alpha_2 = k-2\\x_1 = a+2 & x_2 = k-2
}
$$giving
$$
\frac{3^2}{2+a}+\frac{(k-2)^2}{k-2}\geq \frac{(k+1)^2}{k+a}
$$
Similarly for $b$ in place of $a$, and for $c$ in place of $a$.
Adding the $a$, $b$, and $c$ inequalities we get
$$
\sum_{\mbox{cyc}}\frac1{k+a}\leq \frac1{(k+1)^2}
\sum_{\mbox{cyc}}\left( \frac{3^2}{2+a}+(k-2) \right)
= 3\frac{(k-2)}{(k+1)^2}+\frac{9}{(k+1)^2}\sum_{\mbox{cyc}} \frac{3^2}{2+a}
$$
Now we use our given theorem to write
$$
\sum_{\mbox{cyc}}\frac1{k+a}\leq 3\frac{(k-2)}{(k+1)^2}+\frac{9}{(k+1)^2}=\frac{3k+3}{(k+1)^2}=\frac3{k+1}
$$
