Define $J:\Bbb R^n \to \Bbb R$ by
$ J(y) = \frac{1}{2} (Ay).(y) - b.y, \forall y \in \Bbb R^n $
where $A$ is an $n \times n$ real symmetric and positive definite matrix and $b \in \Bbb R^n$. To show that $J$ is strictly convex we need to show
$\tag{1} J(ty_1 + (1-t)y_2) \lt tJ(y_1) + (1-t)J(y_2) $
So we can evaluate the left:
$ J(ty_1 + (1-t)y_2) = \frac{1}{2} A(ty_1 + (1-t)y_2).(ty_1 + (1-t)y_2) - b.(ty_1 + (1-t)y_2) $
expanding we obtain
$ J(ty_1 + (1-t)y_2) = \frac{1}{2}\left(t^2 Ay_1.y_1 + (1-t)^2Ay_2.y_2+t(1-t)(Ay_1.y_2+Ay_2.y_1)\right) - tby_1-(1-t)by_2 $
From here we can come up with some inequalities, for instance $t^2 \le t$ if $t \in [0,1]$ but there is little I can do.
A possible solution is to take the hessian of $J$ which after some arithmetic comes up to be $A$ and simply say that since $J$ is positive definite (symmetry helps in the computation) we have that $J$ is convex. In this question I'd like to see if there is a way of just showing (1).