# Arithmetic mean - geometric mean [closed]

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'AurizioNov 12 '17 at 17:47

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