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

As hinted by Quasi's solution, repeated squaring works for this problem. This approach should definitely be in your bag, especially since it's so easy to get rid of square roots.   

If you want to simplify it some, then consider a change of variables: $ x = 3a -b , y = 3b-a$.
This gives us $ a = \frac{3x+y}{8}, b = \frac{ x + 3y } { 8}$,   $ a + b = \frac{ x+y} { 2}$ and $ab = \frac{ 3x^2 + 10xy + 3y^2 } { 64}$. 
So, we WTS
$\sqrt{2} \sqrt{ x+y  }  ( \sqrt{x} + \sqrt{y}) \leq \sqrt{ 3x^2 + 10 xy + 3y^2}$
$ \Leftrightarrow 2(x+y) ( x+y + 2 \sqrt{xy} ) \leq 3x^2 + 10xy + 3y^2$
$\Leftrightarrow 4(x+y)\sqrt{xy} \leq x^2 + 6xy + y^2 $
$\Leftrightarrow 0 \leq (\sqrt{x} - \sqrt{y} )^4 $
We have equality iff $ x = y$, or that $ a = b$.   
