# Linear inequality

To prove an algorithm's correctness, I need to show that

$|x|+|x+\Delta h+\Delta s| \geq |x + \Delta s|+|x+\Delta h|$

when $\Delta h > 0$ and $\Delta s > 0$. Mathematica simplifies this to True, but I don't see how. The only tools I know are the triangle inequality and arithmetic/geometric mean inequality.

How can I prove this inequality?

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Have you tried to suppose that $|x|+|x+\Delta h+\Delta s| < |x+\Delta s| + |x+\Delta h |$ and see if this leads to a contradiction? –  Marra Feb 22 '13 at 23:47

In general, any function of the form $|x+a|$ has slope $-1$ to the left of $-a$ and $+1$ to the right of $-a$. This means that $|x+a|+|x+b|$ has slope $-2$ or $0$ or $2$, depending on where $x$ is in relation to $-a$ and $-b$. Now take $a,b$ to be $0,\Delta h+\Delta s$ and $\Delta h,\Delta s$, respectively, and figure out where the corners of the graphs are. (In short, just compute both functions explicitly as piecewise defined by lines, and compare!)

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