3-Variable Constrained Inequality I had a chance to read a newly published problem-solving journal very recently and saw this inequality:
Statement:  If $a, b, c \in [1,\infty)$ with $a+b+c = 6$, then show that:
$\sqrt{a^2-1}+\sqrt{b^2-1} + \sqrt{c^2-1} < \dfrac{3\sqrt{3}abc}{2}$. Just curious about your techniques,etc... 

Let's look at $f(x)= \sqrt{x^2-1}$ for $x\gt1$
$f''(x)=\dfrac {-1}{(x^2-1)^{3/2}}\lt0$ so it's concave and by jensen
$\sqrt{a^2-1}+\sqrt{b^2-1} + \sqrt{c^2-1} \leq 3 \sqrt{(\dfrac {a+b+c}{3})^2-1}=3\sqrt{3}$
By symmetry we can assume $a\geq2 \Rightarrow abc\geq2 \Rightarrow \dfrac{3\sqrt{3}abc}{2}\geq3\sqrt{3}$
(I didn't care equality cases which can be easily eliminated)
