Let $f$ : $\mathbb{R}\rightarrow \mathbb{R}$ be a $C^{2}$ function with $f''(x)>0$ and let
$$F[u(\displaystyle \cdot)]=\int_{0}^{\pi}f(u'(x))dx$$ and $$G[u(\displaystyle \cdot)]=\int_{0}^{\pi}u(x)dx.$$
Now, let $u_*(x)$ be the minimizer of $F[u(\cdot)],$ subject to $G[u(\cdot)]=3$ and $u \in \mathcal{A}$, where $$\mathcal{A}:=\Big\{u:[0,\ 1]\rightarrow \mathbb{R}\ \Big|\ u\in C^{1}[0,1]\text{ and }u(0)=a, u(1)=b\Big\}.$$
Following the usual Calculus of variations optimization methods, I get the Euler-Lagrange equation
$$f''(u_*'(x))u_*''(x) = c_1 \implies u_*''(x) = \frac{c_1}{f''(u_*'(x))},$$ where $c_1$ is an arbitrary constant.
If instead of dividing by $f''(u_*'(x))$, I integrate both sides, then I can get
$$f'(u_*'(x))=c_1x+c_2,$$ but I am not sure how to use any of this to show that $u_*$ is strictly convex. I know that it suffices to show that $u''(x) > 0$, and so, in this case it is enough to show that $c_1 > 0$, but I am not sure how to do that.