Putnam 2018, Problem B-5

Let $f = (f_1, f_2)$ be a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ with continuous partial derivatives $\frac{\partial f_i}{\partial x_j}$ that are positive everywhere. Suppose that
\[
\frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{1}{4} \left( \frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right)^2 > 0
\]
everywhere. Prove that $f$ is one-to-one.

Solution:
Let $(a_1,a_2)$ and $(a_1',a_2')$ be distinct points in $\mathbb{R}^2$; we want to show that $f(a_1,a_2)\neq f(a_1',a_2')$. Write $(v_1,v_2)=(a_1',a_2')-(a_1,a_2)$, and let $\gamma(t)=(a_1,a_2)+t(v_1,v_2)$, $t\in[0,1]$, be the path between $(a_1,a_2)$ and $(a_1',a_2')$. Define a real-valued function $g$ by $g(t)=(v_1,v_2)\cdot f(\gamma(t))$. By the Chain Rule,
$$f'(\gamma(t))=\begin{pmatrix}\partial f_1/\partial x_1&\partial f_1/\partial x_2\\\partial f_2/\partial x_1&\partial f_2/\partial x_2\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}.$$

Abbreviate $\partial f_i/\partial x_j$ by $f_{ij}$; then
$$g'(t)=\begin{pmatrix}v_1&v_2\end{pmatrix}\begin{pmatrix}f_{11}&f_{12}\\f_{21}&f_{22}\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}\\
=f_{11}v_1^2+(f_{12}+f_{21})v_1v_2+f_{22}v_2^2\\
=f_{11}\left(v_1+\frac{f_{12}+f_{21}}{2f_{11}}v_2\right)^2+\frac{4f_{11}f_{22}-(f_{12}+f_{21})^2}{4f_{11}}v_2^2\\
\geq0$$
since $f_{11}$ and $f_{11}f_{22}-(f_{12}+f_{21})^2/4$ are positive by assumption. Since the only way that equality could hold is if $v_1$ and $v_2$ are both 0, we in fact have $g'(t)>0$ for all $t$. But if $f(a_1,a_2)=f(a_1',a_2')$, then $g(0)=g(1)$, a contradiction.