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Let $X$ be a normed space. Show that for every $x,y\in X$: $$\|x+y\|\leq \max\{\|x\|,\|x+2y\|\} $$ Wanted to check two cases. First, assume the maximum is $\|x+2y\|$, then $$\|x+y\|\leq \|x\|+\|y\|\leq\|x+2y\|+\|y\| $$ and stuck. the $y$-norm gets in the way of things.

Makes me think I am either going the wrong way or overlooking something. Hints, please.

AlvinL
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1 Answers1

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\begin{align}\max{\{\|x\|,\|x+2y\|\}}&\overset{(1)}=\frac12\left(\|x\|+\|x+2y\|+\left|\|x\|-\|x+2y\|\right|\right)\\&\overset{(2)}\ge \frac12\|2x+2y\|+\frac12\left|\|x\|-\|x+2y\|\right|\\&\overset{(3)}\ge \|x+y\|\end{align} where $(1)$ holds generally for the maximum of two numbers, $(2)$ is the triangle inequality and $(3)$ is trivial.

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Jimmy R.
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