My analysis:
The second term can be proven to be convex as follows. It is basically a composition of norm with an affine transformation to the power of four:
$(x_1^2+x_2^2+x_3^2+1)^2 = \|(x_1^2, x_2^2, x_3^2, 1)^T\|^4$.
We know that (1) norm is convex (2) composition of norm as a convex function with an affine transformation is convex and (3) raising to the power of an even number, i.e. 4, preserves the convexity of a convex function. Thus, the above term is a composition of a non-decreasing convex function, i.e power of four, with an affine transformation of a norm and thus $(x_1^2+x_2^2+x_3^2+1)^2$ is convex.
The first is term is a challenge for me. I know that $2x_1^2+3x_2^2+x_3^2+4x_1x_2$ can be written as $\mathbf{x}^T A \mathbf{x}$ where
$$ A= \left[ \begin{array}{ccc}
2 & 2 & 0 \\
2 & 3 & 0 \\
0 & 0 & 1
\end{array} \right] $$
It is easy to check that all the eigen values of $A$ are strictly positive which makes $\mathbf{x}^T A \mathbf{x}$ convex. Thus, $2x_1^2+3x_2^2+x_3^2+4x_1x_2+7$ is convex. Now, the issue is that square-root is concave and does not preserve the convexity. How should I proceed from here?
Best Answer
Notice that $\mathrm{x} \mapsto f(\mathrm{x})$ is convex if and only if $y \mapsto f(B\mathrm{y})$ is convex for some invertible matrix $B$. (This is because linear transform preserves lines.)
Now using the spectral theorem, choose a symmetric matrix $B$ such that $AB = BA$ and $A^{-1} = B^2$. Then with the transform $\mathrm{x} = B\mathrm{y}$, we see that the given formula is equal to
$$ (|\mathrm{y}|^2 + 7)^{1/2} + (\mathrm{y}^{T} A^{-1}\mathrm{y} + 1)^2. $$
Now this function is easier to work with, and it is not hard to check that each term is convex. Especially for the first term, following computation may be useful: