<s>[INST] $\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$ I was training for upcoming Olympiads, working on inequalities, and the following inequality came up:
$$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$$ with the obvious delimitations $3b\geq a;\: 3a\geq b.$
I've been pondering the question for quite a while, and tried the using CS among others but didn't find the solution, which, given the level of sophistication the problem should have, surprises me.
Any help would be appreciated.
 [/INST] Presumably we also have the restriction $a,b\ge 0$.

With that assumption we can proceed as follows . . .

If $a+b=0$, then $a=b=0$, and for that case, the inequality clearly holds.

So assume $a+b > 0$.

Since the inequality is homogeneous, the truth of the inequality remains the same if $a,b$ are scaled by an arbitrary positive constant, hence without loss of generality, we can assume $a+b=1$.

Replacing $b$ by $1-a$, it remains to prove
$$
\sqrt{4a-1}+\sqrt{3-4a}\le 4\sqrt{a(1-a)}
\qquad\qquad\;\,
$$
for all $a\in \left[{\large{\frac{1}{4}}},{\large{\frac{3}{4}}}\right]$.

From here it's just routine algebra . . .
\begin{align*}
&
\sqrt{4a-1}+\sqrt{3-4a}\,\le 4\sqrt{a(1-a)}\\[4pt]
\iff\;&
\left(\sqrt{4a-1}+\sqrt{3-4a}\right)^2\le \left(4\sqrt{a(1-a)}\right)^2\\[4pt]
\iff\;&
2+2\sqrt{(4a-1)(3-4a)}\,\le -16a^2+16a\\[4pt]
\iff\;&
\sqrt{(4a-1)(3-4a)}\,\le -8a^2+8a-1\\[4pt]
\iff\;&
(4a-1)(3-4a)\le \left(-8a^2+8a-1\right)^2\\[4pt]
\iff\;&
-16a^2+16a-3\le 64a^4-128a^3+80a^2-16a+1\\[4pt]
\iff\;&
64a^4-128a^3+96a^2-32a+4\ge 0\\[4pt]
\iff\;&
16a^4-32a^3+24a^2-8a+1\ge 0\\[4pt]
\iff\;&
(2a-1)^4\ge 0\\[4pt]
\end{align*}
which is true.

Note:$\;$For the reverse implications, we need to have $-16a^2+16a\ge 0$ and $-8a^2+8a-1\ge 0$, both of which hold since $a\in \left[{\large{\frac{1}{4}}},{\large{\frac{3}{4}}}\right]$.
</s>