Inequality proof: $|f(x)-L| \leq 1 \implies |f(x)| \leq |L| + 1$

Question

I have an inequality proof relating to inequalities for limits in my intro to real analysis class: $$|f(x)-L| \leq 1 \implies |f(x)| \leq |L| + 1$$

Forgive me I know this proof is painfully simple.

My approach was to use the reverse triangle inequality like so.

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Proof:

The reverse triangle inequality gives us $$ |f(x)| - |L| \leq |f(x) - L| \leq 1 $$

adding $|L|$ to all three equations we get

$$ |f(x)| \leq |f(x) - L| + |L| \leq 1 + |L| $$

Therefore $$ |f(x)| \leq |L| + 1 \space \space \space \blacksquare$$

could it be this simple or am I missing something?

Answer 1

Your steps look good. Alternatively observe that $\lvert f(x)-L\rvert \leq 1$ implies $1-L \leq f(x) \leq 1+L$, and $$-\lvert L \rvert - 1 \leq -L-1\leq f(x)\leq 1+L \leq 1+|L|$$ which gives $\lvert f(x) \rvert \leq 1+\lvert L\rvert$. I am not using triangle inequalities here.