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?