Is the exponential of the euclidean norm convex? I know that the euclidean norm is a convex function, and that exponential functions are also convex. I want to know whether or not a mix of the two would also be a convex function. i.e.
$$f(x) = e^{\sqrt{\sum_{i=1}^{n}{x_i^2}}}$$ 
where $$x = (x_1,x_2,...,x_n)$$
So my approach so far has been to prove that 
$$f(\lambda x + (1-\lambda y)) \leq \lambda f(x) + (1-\lambda) f(y)$$
so
$$e^{\lVert \lambda x + (1-\lambda)y\rVert_2}$$ 
but i'm not sure how to proceed from here to get: 
$$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq \lambda e^{\lVert x \rVert_2} + (1-\lambda)e^{\lVert y \rVert_2} $$ 
I've thought about using the triangle inequality but I think that just gets me to: 
$$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq e^{\lambda  \lVert x \rVert_2 + (1-\lambda)\lVert y\rVert_2} = e^{\lambda  \lVert x \rVert_2} + e^{(1-\lambda)\lVert y\rVert_2} $$ 

$$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq e^{\lambda  \lVert x \rVert_2 + (1-\lambda)\lVert y\rVert_2} \leq \lambda e^{  \lVert x \rVert_2} + (1-\lambda)e^{\lVert y\rVert_2} $$ 
Where the first inequality is due to convexity of norm and exponential is an increasing function. The second inequality is due to convexity of exponential function.
Remark: $\exp$ being convex means $\exp(\lambda \alpha +(1-\lambda) \beta) \leq \lambda \exp(\alpha) + (1-\lambda) \exp(\beta)$. $\alpha = \left\| x\right\|, \beta =  \left\| y\right\|$ for your question.
