Integral inequality possibly related to probability theory 
Let $f$ be Riemann integrable such that $\int_a^b f(t) \ dt = 1$ and $f \geq 0$ on $[a,b]$. If $\sigma \in C^2$ is convex, show
$$\sigma\left(\int_a^b tf(t) \ dt\right)\leq\int_a^b f(t)\sigma(t) \ dt $$
and, in addition, discuss when equality holds given the tighter condition $f>0$.

I'm assuming this is related to probability theory, since $f$ meets the conditions to be a probability density function.
I was able to prove this using Jensen's inequality; however, my professor insists that isn't  used. I tried a few other approaches, like fixing one variable and differentiating with respect to the other, but this doesn't seem to lead anywhere. Please help!

Of course it's just Jensen's inequality. Since you don't want to use that, here is a direct proof by integrating by parts several times. 
By integration by parts, $$\int_a^b t f(t) \, \mathrm{d}t = b - \int_a^b F(t) \, \mathrm{d}t$$ where $F(t) = \int_a^t f(s) \, \mathrm{d}x.$
Consider the function $$g(x) = \sigma \Big( x - \int_a^x F(t) \, \mathrm{d}t \Big).$$ It is differentiable with $$g'(x) = (1 - F(x)) \sigma'\Big(x - \int_a^x F(t) \, \mathrm{d}t\Big).$$ Since $\sigma'$ is monotone increasing and $F(t) \ge 0$, we can bound $$g'(x) \le (1 - F(x)) \sigma'(x)$$ and integrating that inequality over $[a,b]$ gives $$\sigma \Big( \int_a^b t f(t) \, \mathrm{d}t \Big) \le \sigma(b) - \int_a^b F(x) \sigma'(x) \, \mathrm{d}x = \int_a^b f(x) \sigma(x) \, \mathrm{d}x,$$ using integration by parts again.
