# Proving absolute value inequality by contradiction

Prove that for $|x|, |y|, |z| \geq 2$ the following holds: $|x^2 + y| + |y^2 + z| + |z^2 + x| \geq |x| + |y| + |z|$

So I thought about a simple proof by contradiction but am not sure whether it's a correct way of thinking for this.

Namely, let's assume $$|x^2 + y| + |y^2 + z| + |z^2 + x| < |x| + |y| + |z|$$ As we know that $|x|, |y|, |z| \geq 2$, the right side is always greater or equal to 6. However, for the case $x=y=z=2$, the left side is also 6 which contradicts the inequality above.

Thus, if $L < R$ isn't true, there has to be a $L \geq R$ correlation between the two q. e. d.

Is that correct?

• No. You only showed $L<R$ is false for a particular case. You'd need to show it's false for every case. – David Mitra Feb 19 '14 at 23:28
• @DavidMitra - yes, you're right. But does that also mean that this inequality can't by proven by contradiction or just the method I thought of is bad? – Straightfw Feb 22 '14 at 13:08

Obviously, the inequality is at its sharpest when $x$, $y$, and $z$, are negative. So in other words, for $x, y, z \geq 2$ we want to show: $|x^2 - y| + |y^2 - z| + |z^2 - x| \geq x + y + z$.
By assumption, $x^2 - 2 x \geq 0$, and similarly for $y$ and $z$.
$x + y + z \leq x + (x^2 - 2x) + y + (y^2 - 2y) + z + (z^2 - 2z) = x^2 - y + y^2 - z + z^2 - x$.