Proving an inequality absolute value

Question

If $g_n \to g$ almost everywhere, then show the following holds almost everywhere:

$$||g_k|^2 - |g - g_k|^2 - |g|^2| \leq 8|g||g - g_k|.$$

By the Triangle Inequality,

\begin{align*} |g_k| &\leq |g| + |g - g_k| \\ |g_k|^2 &\leq |g|^2 + |g - g_k|^2 + 2|g||g - g_k| \\ |g_k|^2 - |g - g_k|^2 - |g|^2 &\leq 2|g||g - g_k| \\ |g_k|^2 - |g - g_k|^2 - |g|^2 &\leq 8|g||g - g_k| \end{align*}

However, I'm not sure how to prove the other way.

Answer 1

I didn't understand "prove other way" fully, is this what you're looking for?

\begin{align*} |g| &\leq |g_k| + |g_k - g| \\ |g| -|g_k - g| &\leq |g_k| \\ |g|^2 +|g_k - g|^2 - 2|g||g - g_k| &\leq |g_k|^2 \\ -2|g||g - g_k| &\leq |g_k|^2 - |g - g_k|^2 - |g|^2 \\ -8|g||g - g_k| &\leq |g_k|^2 - |g - g_k|^2 - |g|^2 \\ \end{align*}