A brief proof of Convexity. Given $f(x,y) = \sqrt{1 + x^2 + y^2}$, prove that $f(x,y)$ is convex. Try to do this without using the Hessian.
 A: Let $z = (x,y)$,
$$f(z)=\left\| \begin{array}{c} z \\ 1\end{array}\right\|$$
\begin{align}f(\lambda z_1 + (1-\lambda) z_2) &= \left\| \begin{array}{c} \lambda z_1 + (1-\lambda)z_2 \\ 1\end{array}\right\| \\
&= \left\| \begin{array}{c} \lambda z_1 + (1-\lambda)z_2 \\ \lambda+(1-\lambda)\end{array}\right\| \\&\leq \lambda f(z_1)+(1-\lambda)f(z_2)\end{align}
by triangle inequality.
A: We can use the definition of convexity directly, i.e., to prove that 
$$f\big(\lambda(x_1,y_1) + (1 - \lambda)(x_2, y_2)\big) \leq \lambda ~f(x_1,y_1) + (1-\lambda)~f(x_2, y_2)$$
Based on the basic algebraic calculation, plus the note that:
$$ 2x_1x_2y_1y_2 \leq x_1^2y_2^2 + x_2^2y_1^2 $$ 
A: Since $f$ is clearly continuous, in order to prove convexity it is enough to show the midpoint-convexity, i.e. the inequality
$$\sqrt{1+a^2+b^2}+\sqrt{1+c^2+d^2}\geq \sqrt{4+\left(a+c\right)^2+\left(b+d\right)^2} \tag{1}$$
that boils down to
$$\sqrt{(1+a^2+b^2)(1+c^2+d^2)} \geq 1+ac+bd \tag{2} $$
that is a straightforward consequence of the Cauchy-Schwarz inequality.
A: Graphically. The contour curves: $$x^2+y^2=z^2-1$$ are circles with the radius $\sqrt{z^2-1}$ :

As the lower contour set is convex, $z$ is a convex function.
