Use arithmetic mean -geometric mean inequality to prove that perimeter of a rectangle is minimum with a given area if it is a square


closed as off-topic by user223391, Jack D'Aurizio Nov 12 '17 at 17:47

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    $\begingroup$ That pretty much is the AM/GM inequality (for two variables). $\endgroup$ – Lord Shark the Unknown Nov 12 '17 at 17:19
  • $\begingroup$ I don't see a question. $\endgroup$ – user223391 Nov 12 '17 at 17:23

Let $a$ and $b$ be two adjacent sides, $P$ be the perimeter and $S$ be an area of the rectangle.

Thus, by AM-GM $$P=2(a+b)=4\cdot\frac{a+b}{2}\geq4\sqrt{ab}=4\sqrt{S}.$$ The equality occurs for $a=b$, id est, when our rectangle is a square.


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