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So far I have the first case when $a=b$: \begin{align*} |ab| &= |b^2|\\ &=|b|^2\\ &=\frac{2|b|^2}2\\ &=\frac{b^2+b^2}2\\ &=\frac{a^2+b^2}2 \end{align*}

Case 2: $a>b$

Case 3 $a<b$

I've been stuck on this problem for a few hours now and don't know how to proceed. Am I approaching it wrong? Should I not be thinking about the cases of $a$ in terms of $b$? Thank you for your help.

dfeuer
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david stocker
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2 Answers2

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Without loss of generality we may assume that $a,b\ge 0$. Rewrite the inequality as $2ab\le a^2+b^2$, and then as $(a-b)^2\ge 0$.

Or else, start from the obvious fact that $(|a|-|b|)^2\ge 0$. Expanding, rewrite this as $2|ab|\le a^2+b^2$, and it's almost over.

André Nicolas
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  • the second part makes the most sense. the first part assumes that the inequality is true already. Thank you for the help. it was a life saver – david stocker Oct 22 '13 at 02:23
  • You are welcome. For the first argument, one one write (for non-negative $a,b$) that $ab\le \frac{1}{2}(a^2+b^2)$ if and only if $2ab\le a^2+b^2$ if and only if $(a-b)^2\ge 0$. A lot of inequalities ultimately come down to the fact that a square is always non-negative. – André Nicolas Oct 22 '13 at 02:28
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If $ab\ge0$ then $|ab|=ab\le ab+\dfrac{(a-b)^2}2=\dfrac{a^2+b^2}2$.
If $ab\le0$ then $|ab|=-ab\le-ab+\dfrac{(a+b)^2}2=\dfrac{a^2+b^2}2$.

bof
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  • this is the format my professor would prefer. starting at the premise and ending with the conclusion with all cases included. – david stocker Oct 22 '13 at 03:04