Need help to decide if $\{x \in \mathbb R^n : 1 \leq x_1^2 + x_2^2 + \cdots + x_n^2 \leq 2 \}$ is convex 
Is the following set convex? $$\{x \in \mathbb R^n : 1 \leq x_1^2 + x_2^2 +  \cdots + x_n^2 \leq 2 \}$$

I did the following. 
Assume $1≤x_1^2+x_2^2+...+x_n^2≤2$ and $1≤y_1^2+y_2^2+...+y_n^2≤2$
Assume $z=αx+(1-α)y, α∈[0,1]$
Now I have to prove that $1≤z_1^2+z_2^2+...+z_n^2≤2$
Solving the right hand part is relatively easy as in the end I get $2α^2 + 2(1-α)^2 + 2α(1-α)(x_1y_1+...+x_ny_n)$ which is greater than $2α^2 + 2(1-α)^2 + 2α(1-α)(\sqrt{x_1^2+x_2^2+...+x_n^2} *\sqrt{y_1^2+y_2^2+...+y_n^2})$ according to Cauchy-Schwartz. In the end it comes down to $2(α+(1-α))^2$ which is 2. 
On the left side, however I cannot use Cauchy-Schwartz because it decreases the value and I need to show that my equation of $z_1^2+z_2^2+...+z_n^2≥1$
Does it mean that the set is not convex? 

The set is not convex because $(\pm1,0,0,\ldots,0)$ belong to it, but $(0,0,0,\ldots,0)$ doesn't. Note that$$(0,0,0,\ldots,0)=\frac12(1,0,0,\ldots,0)+\frac12(-1,0,0,\ldots,0).$$
