I am trying to solve the following exercise:
Let $F:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a strictly convex function which is $C^2$, non-negative and such that $F(0,0)=0$.
1). Prove that the equation $F(x,y)=F(0,1)$ implicitly defines a function $y=\phi(x)$ in $C^2$ for $x$ around $0$ such that $\phi(0)=1$;
2). Prove that $\phi''(0)\leq 0$.
I have proved the first point by using the Dini's theorem but I can't solve the second point by using the fact that $\phi'(x)=-\frac{F_x(x,\phi(x))}{F_y(x,\phi(x))}$.
We have
$$F(x,\phi(x))=F(0,1)$$
Differentiate both sides with respect to $x$:
$$F_x+F_y\phi'=0$$
Differentiate again
$$F_{xx}+F_{xy}\phi'+(F_{yx}+F_{yy}\phi')\phi'+F_y\phi''=0$$
or
$$-F_y\phi''=F_{xx}+2F_{xy}\phi'+F_{yy}(\phi')^2.$$
Substitute for $\phi'$:
$$-F_y\phi''=F_{xx}-2F_{xy}\frac{F_x}{F_y}+F_{yy}\frac{F_x^2}{F_y^2}.$$
Simplify
$$-F_y^3\phi''=F_y^2F_{xx}+F_x^2F_{yy}-2F_xF_yF_{xy}$$
If $F_{xy}(0,1)=0$, then we get the desired result because convexity of $F$ implies $F_{xx}\geq 0$ and $F_{yy}\geq 0$. So suppose $F_{xy}(0,1)\neq 0$. Now by convexity of $F$ we know:
$$F_{xx}F_{yy}\geq F_{xy}^2(>0)\iff F_{yy}\geq \frac{F_{xy}^2}{F_{xx}}$$
Thus
$$-F_y^3\phi''\geq F_y^2F_{xx}+F_x^2\frac{F_{xy}^2}{F_{xx}}-2F_xF_yF_{xy}$$
or, simplifying,
$$-F_y^3\phi''\geq \frac{1}{F_{xx}}[F_y^2F_{xx}^2+F_x^2F_{xy}^2-2F_xF_yF_{xx}F_{xy}]=\frac{1}{F_{xx}}(F_yF_{xx}-F_xF_{xy})^2.$$
Since $F_{xx}(0,1)>0$ by convexity of $F$, the right hand side is nonnegative. Also $F_y(0,1)$ is positive because $F(0,1)>0=F(0,0)$ (if not $F$ would not be strictly convex). The result follows.