Show that if $\varphi$ is strictly convex and $\varphi(EX)=E(\varphi(X))$ then $X=EX$ almost surely 
Show that if $\varphi$ is strictly convex and $\varphi(EX)=E(\varphi(X))$ then $X=EX$ almost surely.

Here we assume that both $EX < \infty$ and $E( \varphi (X)) < \infty$
My solution:
Now since $\varphi$ is strictly convex we have that  $$\varphi (EX) <E(\varphi(X) )$$ and from the problem statement we have that  $\varphi(EX)=E(\varphi (X))$
Now the above two statements together are false and since false implies any statement whether its true or false, I have the required result
Does this make any sense?
My reasoning looks a bit incongruous and nonsensical but I can't think of anything else.
Is it correct?
How could I show what needs to be shown otherwise
 A: By strict convexity there is a subdifferential $m$ of $\varphi$ at $E[X]$ so that
$$\varphi(x) > m (x - E[X]) + \varphi(E[X])$$
holds for all $x \ne E[X]$. Note the following equality:
$$E[\varphi(X) - m(X - E[X]) - \varphi(E[X])] = E[\varphi(X)] - \varphi(E[X]) = 0$$
Since the integrand $\varphi(X) - m(X - E[X]) - \varphi(E[X])$ is nonnegative, the only way that its expectation can be zero is that the integrand itself is almost surely zero. By the preceding inequality this can only hold if $X = E[X]$ almost surely.
A: Let $t = \Pr(X \ge EX)$.
Then, we have
$$ EX = (1-t)E[X|X < EX] + tE[X|X \ge EX] $$
and
\begin{align} 
\varphi(EX) 
&= \varphi((1-t)E[X \mid X < EX] + tE[X \mid X \ge EX]) \\
&\le (1-t)\varphi(E[X \mid X < EX]) + t\varphi(E[X \mid X \ge EX]) \\
&\le (1-t)E[\varphi(X) \mid X < EX] + tE[\varphi(X) \mid X \ge EX] \\
&=E[\varphi(X)].
\end{align}
Now, by assumption we also have $\varphi(EX) = E\varphi(X)$. From the strictly convexity of $\varphi$ follows 1) $t=1$, or 2) $t=0$, or 3) $E[X|X<EX] = E[X|X\ge EX]$.
Case 1: If $t=1$, that is $X - EX \ge 0$ a.s. Then, we have
$$ 0 \le E[X - EX] = EX - EX = 0. $$
Thus, $X = EX$ a.s.
Case 2: If $t = 0$, then we have $EX - X > 0$ a.s. and 
$$ 0\le E[EX - X] = EX - EX = 0, $$
which implies $X = EX$ a.s, a contradiction. 
Thus, this case is not possible.
Case 3: Without loss of generality assume $t < 1$. Assume $E[X \mid X<EX] = E[X \mid X\ge EX]$. Thus $E[X \mid X<EX] = E[X \mid X\ge EX] = EX$. But again, we have
$$ 0 \le E[EX - X \mid X \le EX] = EX - E[X \mid X \le EX] \le EX - EX = 0. $$
That is $\Pr(X=EX \mid X \le EX) = 1$ and 
$$ \Pr(X=EX) = \Pr(X=EX , X \le EX) = \Pr(X \le EX), $$
which implies $\Pr(X < EX) = 0$ and $\Pr(X \ge EX) = 1 > t$.
That is, this case is also not possible.
