Real Numbers Inequality Proof: x+y+z > (|x|+|y|+|z|)/3

[anemone]

Here is this week's POTW:

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Suppose $x,\,y,\,z$ are real numbers such that $x+y>0$, $y+z>0$ and $z+x>0$.

Prove that $x+y+z>\dfrac{|x|+|y|+|z|}{3}$.
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[anemone]

Congratulations to the following members for their correct solutions::)

1. lfdahl
2. greg1313

Solution from greg1313:

Spoiler

If x > 0, y > 0, z > 0 we are done.

From the given inequalities only one of x, y, z can be non-positive and by symmetry this can be anyone of x, y, z; the outcome will be the same.

Choosing y to be non-positive, we also have |x| > |y| and |z| > |y|.

Now, with y non-positive

$$x + y + z = |x| - |y| + |z| = \frac{3|x|-3|y|+3|z|}{3}$$

If the difference $$\frac{3|x|-3|y|+3|z|}{3}-\frac{|x|+|y|+|z|}{3}$$ is positive we are done.

$$\frac{3|x|-3|y|+3|z|}{3}-\frac{|x|+|y|+|z|}{3}=\frac{2(|x|+|z|)-4|y|}{3}$$

but since |x| > |y| and |z| > |y|, |x| + |z| > 2|y| so $$\frac{2(|x|+|z|)-4|y|}{3}>0$$ as required.