Is $\|x_1 - 2x_2\| \le \sqrt{5} \|x\|$ true?

Question

Is the following identity true? If so, how?

$$\|x_1-2x_2\| \le \sqrt{5} \|x\|$$

where $x := (x_1,x_2)$.

Answer 1

For a direct proof, working backwards, we have

$$(2x_1+x_2)^2\geqslant 0 \iff 4x_1^2+4x_1 x_2+x_2^2\geqslant 0 $$

$$\iff (x_1-2x_2)^2\leqslant 5(x_1^2+x_2^2)$$

$$\iff |x_1-2x_2|\leqslant \sqrt{5} ||x||_2.$$

We can also use the Cauchy-Schwarz inequality: notice that for $y=(1,-2)$, we have $$|x_1-2x_2|=|\langle x,y\rangle|\leqslant ||x|| ||y||=\sqrt{5}||x||.$$