Can we use concavity to prove the following statement
$$f(\theta a + (1-\theta)c, \theta b + (1-\theta)d) \geq \theta f(a, b) + (1-\theta) f(c, d)$$ The $f(x, y)$ can be defined as $f(x, y) = x^{1\over 2}y^{1\over 2}$. Or use other method to prove this. Thanks!
The inequality you have written is true for concave functions (it is basically Jensen's inequality for -f). The function is twice differentiable on $(0,\infty)^2$, so we may use its Hessian matrix. A twice-differentiable function is concave if and only if its Hessian matrix is negative-semidefinite (has all eigenvalues negative or zero).
\begin{align} H &= \begin{pmatrix}-\frac{\sqrt{y}}{4x^\frac{3}{2}} & \frac{1}{4\sqrt{x}\sqrt{y}}\\ \frac{1}{4\sqrt{x}\sqrt{y}}& -\frac{\sqrt{x}}{4y^\frac{3}{2}}\end{pmatrix}\\ &= \frac{1}{4\sqrt{x}\sqrt{y}}\begin{pmatrix}-\frac{y}{x} & 1\\ 1& -\frac{x}{y}\end{pmatrix} \end{align} We can compute the eigenvalues of the second matrix as solutions of \begin{align} \left(\lambda + \frac{y}{x}\right)\left(\lambda + \frac{x}{y}\right) -1 = 0 \end{align} this is pretty easy to solve with the quadratic formula \begin{align} 0&=\lambda^2 + \lambda\left(\frac{y}{x} + \frac{x}{y}\right) + 1 -1\\ \lambda_\pm &= -\frac{1}{2}\left(\frac{y}{x} + \frac{x}{y}\right)\pm\frac{1}{2}\sqrt{\left(\frac{y}{x} + \frac{x}{y}\right)^2 }\\ \implies \lambda_+ &= 0\\ \lambda_- &= -\frac{y}{x} - \frac{x}{y}. \end{align}
So one eigenvalue is negative and the other is zero, so the matrix is negative-semidefinite and $f$ is concave.
