\(|a+b| \leq |a|+|b|\) holds for all real numbers \(a\) and \(b\).

For the absolute value function definition and other properties see here.

Proof:

Let \(a\) and \(b\) be real numbers. Expand \((|a| + |b|)^2\) to see

$$ \begin{align*} (|a| + |b|)^2 &= |a|^2 + 2|a||b| + |b|^2 \\ &= a^2 + 2|a||b| + b^2 \quad (\text{because \(|a|^2 = a^2\)})\\ &\geq a^2 + 2ab + b^2 \quad (\text{because \(|a| \geq a\) why? (1) below})\\ &= (a + b)^2 \\ &= |a + b|^2 \quad (\text{because \(|x|^2 = x^2\)}) \end{align*} $$

So now we have

$$ \begin{align*} (|a| + |b|)^2 &\geq |a + b|^2 \\ |a| + |b| &\geq |a+b| \end{align*} $$

as required. \(\blacksquare\)
(1) I was unable to add the proof link inside the math formula but for the proof of \(|a| \geq a\), see see here.

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