prove strictly convexity I wanna show $f(x)=-x^{\top}p+c\cdot\sqrt{x^\top\Sigma x}$ is strictly convex, where $p$ is a constant vector, $\Sigma$ is positive definite. I tried to show it by using $f''(x)$ is positive definite and I could obtain
$$H(x)=c\cdot \frac{\Sigma -\frac{(\Sigma x)(\Sigma x)^{\top}}{x^\top\Sigma x}}{\sqrt{x^\top \Sigma x}}$$
yet I am stuck here. Could someone help me clarify it?
Thanks in advance!
 A: Your function is not differentiable in $x = 0$, so you can't prove strict convexity by using the second derivative.
Also, the function is not strictly convex. Note that $f$ is positively homogenous of degree $1$, which implies that for any $x \ne 0$ the following holds: 
$$f\big(\tfrac{1}{2} x + \tfrac{1}{2}2x\big) = f\big(\tfrac{3}{2} x) = \frac{3}{2} f(x) = \frac{1}{2} f(x) + \frac{1}{2} 2f(x) = \frac{1}{2}f(x) + \frac{1}{2} f(2x).$$
However, the function is indeed convex for $c \ge 0$. We can ignore the part $-x^tp$, since this is a linear function; The only difficult part is proving that $x \mapsto \sqrt{x^t \Sigma x}$ is convex. To this extent, note that for any $x, y$ and $\lambda \in [0, 1]$ the following holds by the Cauchy-Schwarz inequality, applied to the inner product $\langle x, y \rangle = x^t \Sigma y$:
$$(\lambda x + (1 - \lambda)y)^t\Sigma (\lambda x + (1 - \lambda y) = \lambda^2 x^t \Sigma x + 2(1 - \lambda) \lambda x^t \Sigma y + (1 - \lambda)^2 y^t \Sigma y\\
\le \lambda^2 x^t \Sigma x + 2 (1 - \lambda) \lambda \sqrt{x^t \Sigma x} \sqrt{y^t \Sigma y} + (1 - \lambda)^2 y^t \Sigma y = \big(\lambda \sqrt{x^t\Sigma x} + (1 - \lambda) \sqrt{y^t \Sigma y}\big)^2$$
Note that in the first equality we have used that $\Sigma$ is wlog symmetric, which implies 
$$y^t\Sigma x = (y^t \Sigma x)^t = x^t \Sigma^t y = x^t \Sigma y.$$
