The definition of strongly convex from Wikipedia:
It is not necessary for a function to be differentiable in order to be strongly convex. A third definition for a strongly convex function, with parameter $m$, is that, for all $x$, $y$ in the domain and $t\in [0,1]$, $f(tx+(1-t)y) \le t f(x)+(1-t)f(y) - \frac{1}{2} m t(1-t) \|x-y\|_2^2.$
Prove that the 2-norm squared $f(w) = m\|w\|^2 $ is m strongly convex
I have so far tried to use the triangle inequality but I cannot derive that last negative term.