We know $a \geq a' \geq 0$ and $b\geq b' \geq 0$ . How we can prove:


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    $\begingroup$ Do it by cases: either the $\max$ is $a'$ or it is $b'$. Discuss each case separately. Remember that if $a'=\max \left(\left\{a'b'\right\}\right)$, then $b'\leq a'$ and similarly if $b$ is the $\max$. $\endgroup$ – Git Gud Nov 13 '13 at 14:52
  • $\begingroup$ You're absolutely right. It was really simple. :) $\endgroup$ – 2012User Nov 13 '13 at 14:56

Hint: Assuming that $a',b'>0$, use the fact that $\max\{a',b'\} \ge a',\ \max\{a',b'\}\ge b'$.


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