$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$ 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.

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 $a\in \left[{\large{\frac{1}{4}}},{\large{\frac{3}{4}}}\right]$.
