I've been reflecting on the following problem: let $f \in \mathcal{C}^3([a,b] \times \mathbb{R} \times \mathbb{R})$ and $$ (P) \quad \inf_{u \in X} \left \{ I(u)=\int_{a}^{b} f(x,u(x),u'(x)) \, dx\right \} $$
where $X=\left \{ u \in \mathcal{C}^1([a,b]):u(a)= \alpha, u(b)= \beta \right \}$. Let $\bar{u} \in X \cap \mathcal{C}^2([a,b])$ be a minimizer for $(P)$. Show that the following inequality $$ \int_{a}^{b} [f_{uu}(x,\bar{u},\bar{u}')v^2+2f_{u \xi}(x,\bar{u},\bar{u}')vv'+f_{\xi \xi}(x,\bar{u},\bar{u}')v'^2] \,dx \geqslant 0 $$ holds for every $v \in \mathcal{C}^1_0(a,b)$.
If $\bar{u}$ is a minimizer for $(P)$ then $\bar{u}$ satisfies the Euler equation $$ \frac{d}{dx}[f_{\xi}(x,\bar{u},\bar{u}')]=f_{u}(x,\bar{u},\bar{u}'), \quad x \in (a,b), $$ then, setting $$ \phi(h)=I(\bar{u}+hv)=\int_{a}^{b} f(x,\bar{u}+hv,\bar{u}'+hv') \, dx $$ we have that $\phi(0) \leqslant \phi(h)$ for every $h \in \mathbb{R}$, then $\phi'(0)=0$. Considering that $\bar{u}$ is a minimizer we would have that $\phi(h)$ is concave up at $h=0$, then $\phi''(0) \geqslant 0$, and $\phi''(0)=\int_{a}^{b} [f_{uu}(x,\bar{u},\bar{u}')v^2+2f_{u \xi}(x,\bar{u},\bar{u}')vv'+f_{\xi \xi}(x,\bar{u},\bar{u}')v'^2] \,dx$.
My question is if this solution is right or if I've made a mistake.