I'm having trouble proving that a function, $f(x,y)=\left(a\sqrt{x}+b\sqrt{y}\right)^2$, is concave for $\forall x, y \in (0,\infty)$ where $a > 0$ and $b > 0$ using the definition for concavity: $$f(\lambda x + (1 - \lambda )y) \geq \lambda f(x) + (1 - \lambda )f(y),\ \forall\lambda\in [0,1]$$ Here's what I have so far: \begin{array}{rcl} f(\lambda x_1 + (1 - \lambda) x_2,\,\lambda y_1 + (1 - \lambda) y_2) &=& \left(a\sqrt{\lambda x_1+(1-\lambda)x_2}+b\sqrt{\lambda y_1+(1-\lambda)y_2}\right)^2 \\ &=& a^2(\lambda x_1+(1-\lambda)x_2)+b^2(\lambda y_1+(1-\lambda)y_2)\\ &&{}+2ab\sqrt{\lambda x_1+(1-\lambda) x_2}\sqrt{\lambda y_1+(1-\lambda) y_2} \end{array}
I'm aware that using the second derivatives is much less hassle, but I'd prefer using the definition (if at all possible) as the problem didn't specify how to solve it. Any inputs would be much appreciated!