$3$-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?
 A: We need to prove that $$\sum_{cyc}\frac{a^2b^2}{3+c}\leq\frac{3}{4},$$ where $a$, $b$ and $c$ are positives such that $a^2+b^2+c^2=3$.
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition does not depend on $w^3$ and we need to prove that
$$4\sum_{cyc}a^2b^2(a+3)(b+3)\leq3\prod_{cyc}(a+3)$$ and since
$$\sum_{cyc}a^3b^3=27v^6-27uv^2w^3+3w^6$$ and the rest terms are less degree, we need to prove that $f(w^3)\geq0,$ where $f$ is a concave function.
But a concave function gets a minimal value for an extremal value of $w^3$, which by $uvw$(see here: https://artofproblemsolving.com/community/c6h278791) happens in the following cases.
1.$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$.
Thus, we need to prove that $$\frac{a^2b^2}{3}\leq\frac{3}{4}$$ or $$(a^2-b^2)^2\geq0.$$
2.Two variables are equal.
Let $b=a$. Thus, $c=\sqrt{3-2a^2},$ where $0<a<\sqrt{\frac{3}{2}}$ and we need to prove that
$$\frac{a^4}{3+\sqrt{3-2a^2}}+\frac{2a^2(3-2a^2)}{3+a}\leq\frac{3}{4}.$$
Can you end it now?
A: 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.
