$3$-var inequality: $\frac{bc}{\sqrt{a}+3}+\frac{ca}{\sqrt{b}+3}+\frac{ab}{\sqrt{c}+3} \leq \frac{3}{4}$ for $a+b+c=3$. Problem: Let $a,b,c$ be positive numbers satisfied $a+b+c=3$. Prove that $$\dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \leq \dfrac{3}{4}$$
I've tried U.C.T method but it doesn't reach the solution. I also thought of $p,q,r$ method, but I think it will end up a ugly solution. The only useful thing I get is $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca$. Anyone?

Fact 1: It holds that, for all $x \ge 0$,
$$\frac{1}{3 + \sqrt x} \le \frac13 - \frac{13}{96}x + \frac{5}{96}x^2.$$
(Note: Letting $x=y^2$,
we have $\mathrm{RHS} - \mathrm{LHS} = \frac{y(5y^2 + 25y + 32)(y-1)^2}{96(y+3)}\ge 0$.
The RHS comes from the Taylor approximation of the LHS around $x=1$,
that is $\frac{1}{3 + \sqrt x}\sim \frac14 - \frac{1}{32}(x-1) + \frac{3}{256}(x-1)^2$, after adjusting the coefficient of $(x-1)^2$.)
By Fact 1, it suffices to prove that
$$\sum_{\mathrm{cyc}} bc \left(\frac13 - \frac{13}{96}a + \frac{5}{96}a^2\right) \le \frac34$$
or
$$\frac13(ab + bc + ca) - \frac{13}{32}abc + \frac{5}{96}abc(a + b + c) \le \frac34$$
or (using $a + b + c = 3$)
$$\frac13(ab + bc + ca) - \frac{1}{4}abc \le \frac34$$
or
$$12(ab + bc + ca) - 9abc \le 27$$
or (using $a + b + c = 3$)
$$4(a + b + c)(ab + bc + ca) - 9abc \le (a + b + c)^3$$
which is degree three Schur.
We are done.
