<s>[INST] $3$-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?
 [/INST] 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?
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