Prove or disprove $|a| \geq 2|b| \implies |a+b| \geq |b|$

Question

I conjectured that for all real $a, b$, that $|a| \geq 2|b| \implies |a+b| \geq |b|$. I'm using this to try to prove a limit where the denominator is $x^2+xy+y^2$ since I know $|x^2+y^2| \geq 2|xy|$.

My attempt: Suppose $|a| \geq 2|b|$. We want to show that $|a+b| \geq |b|$.

$|a| \geq 2|b| \iff |a|-|b| \geq |b| $. Since $|b| \geq 0$ then $|a|+|b| \geq |a|-|b|$.

So we have $|a| + |b| \geq |b| \geq 0.$ Therefore $|a| + |b| = ||a| + |b||$. But I don't know whether $||a|+|b|| \leq |a+b|$. How to proceed, or is my conjecture wrong? And if it is wrong, is there any tweak that can be made to make it correct?

Also, what would be an alternative way to minorate $|x^2+xy+y^2|$ ?

Answer 1

Hint:

By the reverse triangle inequality, you have,

$$|a-b|\geq ||a|-|b||~\forall~a,b\in\Bbb C$$

Replacing $b$ by $-b$ and using the given relation $|a|\geq 2|b|$ gives you the desired inequality.

Answer 2

Suppose $|a| \geq 2|b|$. Then $|a|-|b| \geq |b|$, so in particular, $||a|-|b|| \geq |b|$. By the triangle inequality you get $$|a+b| \geq \left||a|-|b| \right| \geq |b|$$

Answer 3

If $b=0$ this is true. Say $x=a/b$ then we have $|x|\geq 2$ and we have to prove $|x+1|\geq 1$. So $$x\geq 2 \implies x+1\geq 3\implies |x+1|>1$$ or $$x\leq -2\implies x+1 \leq -1 \implies |x+1|\geq 1$$