Let us discuss about 'inequality' related problems - Minimum Perimeter Problem. All algebraic inequality problems can be traced back to two key ideas:
Though these two notions seem trivial and obvious in nature, they lead to a very rich and diverse theory of algebraic inequality. We will examine some of these ideas.
Consider all rectangles whose area is 1. What is the minimum perimeter of such rectangles?
Clearly maximum perimeter can be as large as you please. For example, you may take one side's length to be 1,00,000 and the other side's length as $latex \frac{1}{1,00,000} $. Then the area is $latex 1,00,000 \times \frac{1}{1,00,000} = 1 $ but the perimeter is $latex 1,00,000 + \frac{1}{1,00,000} + 1,00,000 + \frac{1}{1,00,000} $ (which is a large number).
Using similar technique one may make the perimeter as large as he wishes, keeping the area fixed (=1). In some sense the rectangle will become thinner in the process.
Hence, more critical question is: what is the smallest possible perimeter (or whether the perimeter can be made as small as we wish)?
Algebraically the problem can be designed as follows:
If a is the length of a rectangle and $latex \frac{1}{a} $ is the width (thus area $latex a \times \frac{1}{a} = 1 $ ) then what is the minimum value of the perimeter $latex 2\left (a + \frac{1}{a} \right ) $
Since 2 is a constant, we may search for the minimum value of $latex a + \frac{1}{a} $.
There are several

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