Algebraic approach towards convexity I have a function: $ u(x) = x_{1} + x_{2} + \min\{x_{1}, x_{2}\}$. How do we algebraically show if it's convex or not? Also, what would be the general way to show if any given function is convex.
I assume you meant $u(x_1,x_2)=x_1+x_2+\min\{x_1,x_2\}$. To check if a function of two or more variables is convex, you need to check if you can differentiate it twice, calculate its Hessian and check if it is positive semi-definite. On top of that the domain needs to be a convex set but since we talk about utility in economics - you don't need to worry about that too much - Can A Utility Function Take On Negative Values?

In your example: 
$$\mathbf{H}=\begin{bmatrix}
 \frac{\partial^2u}{\partial x_1^2} &amp; \frac{\partial^2u}{\partial x_1x_2}  \\
 \frac{\partial^2u}{\partial x_1x_2} &amp; \frac{\partial^2u}{\partial x_2^2}  \\
\end{bmatrix}=\begin{bmatrix}
 0 &amp; 0  \\
 0 &amp; 0  \\
\end{bmatrix}$$

so it is?

The approach described above is a general way of showing a function being convex but as pointed in the comment section, you need to be careful - make sure that all the assumptions are met before jumping to conclusions. (Or else, you end up being mistaken as me right now.)

EDIT.

Minimum function mislead me, I agree. 

Let's get back to basics:

$u(x_1,x_2)=x_1+x_2+\min(x_1,x_2)=x_1+x_2 + \frac{x_1+x_2-|x_1-x_2|}{2}=\begin{cases}x_1+2x_2 \;\; \mathrm{for} \;\; x_1 \geq x_2\\ 2x_1+x_2 \;\; \mathrm{for} \;\; x_1&lt;x_2 \end{cases}$ 

It looks like that (on horizontal axis $x_1$, on vertical axis $x_2$: 


It is neither convex, nor concave. Due to the absolute value function that is in a core of minimum function: 
Why is the absolute value function not differentiable at x=0?

I thought that eliminating point where $x_1=x_2$ would solve the issue but as shown in the other answer - it does not.

EDIT. 2 (final)

It is, in fact, concave as the user @afreelunch was telling me. You can see it directly from the definition. Or (what I think is even easier) from the way our utility function is built. 

We have $u(x_1,x_2) = \min\{2x_1+x_2,x_1+2x_2\}$. Now, let's take it apart: $f_1(x_1) = 2x_1$ and $g_1(x_2) = x_2$, they are both linear Is linear function convex or concave? - hence, concave (and convex but we care about concavity here) and therefore their sum (property of concave function) $h_1(x_1,x_2)=2x_1+x_2$ is also concave. Similarly, $f_2(x_1) = x_1$ and $g_2(x_2)=2x_2$ and $h_2(x_1,x_2)=x_1+2x_2$ is concave as well. Knowing that, we get $\min\{h_1(x_1,x_2),h_2(x_1,x_2)\}$ - minimum of two concave functions - which is then concave (also a well known fact). 

In short, minimum preserves concavity (and maximum preserves convexity).