$\displaystyle - \sqrt{a ^{2}+b ^{2} } \leqslant a \cdot cos \alpha + b \cdot sin \alpha \leqslant \sqrt{a ^{2}+b ^{2} } $

My solution:

$\displaystyle \begin{cases} - \sqrt{ a^{2}+b^{2} } \leqslant a \cdot cos \alpha +b \cdot sin \alpha \\a \cdot cos \alpha +b \cdot sin \alpha \leqslant \sqrt{ a^{2}+b^{2} }\end{cases}$

After adding up both equalities we have:

$\displaystyle a \cdot cos \alpha +b \cdot sin \alpha-\sqrt{ a^{2}+b^{2} } \leqslant \sqrt{ a^{2}+b^{2} } + a \cdot cos \alpha +b \cdot sin \alpha$

$\displaystyle 2\sqrt{ a^{2}+b^{2} } \geqslant 0$

My question is as follows:

Is this solution correct, or is it after adding up the two equalities still obvious to write two?

I hope you understand what I mean :]