How to prove the concavity of this function?

Question

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!

Answer 1

\begin{equation} \begin{aligned} \lambda f(x_1,y_1) + (1-\lambda) f(x_2,y_2) &= \lambda(a^2x_1+b^2y_1 + 2ab\sqrt{x_1y_1}) + (1-\lambda) (a^2x_2 +b^2y_2 + 2ab\sqrt{x_2 y_2}) \\ &= \lambda(a^2x_1+b^2y_1)+(1-\lambda)(a^2x_2+b^2y_2) + 2ab(\lambda \sqrt{x_1y_1}+(1-\lambda)\sqrt{x_2y_2}). \end{aligned} \end{equation} Due to the $g(x,y)=\sqrt{xy}$ is concave on $(x,y) \in \mathbb{R}^2_{+}$, thus \begin{equation} \lambda f(x_1,y_1) + (1-\lambda) f(x_2,y_2) \leq \lambda(a^2x_1+b^2y_1)+(1-\lambda)(a^2x_2+b^2y_2) + 2ab\sqrt{(\lambda x_1+ (1-\lambda x_2)(\lambda y_1 + (1-\lambda y_2)))} \end{equation} Thus $f(x,y)$ is also concave on its domain.