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Algebra (Advanced) - AM and GM

Today, let’s look in detail at a relation between arithmetic mean and geometric mean of two numbers. It is one of those properties which make sense the moment someone explains to us but are very hard to arrive on our own.


When two positive numbers are equal, their Arithmetic Mean = Geometric Mean = The number itself


Say, the two numbers are m and n (and are equal). Their arithmetic mean = (m+n)/2 = 2m/2 = m


Their geometric mean = sqrt(m*n) = sqrt(m^2) = m (the numbers are positive so |m| = m)


We also know that Arithmetic Mean >= Geometric Mean


So when arithmetic mean is equal to geometric mean, it means the arithmetic mean is taking its minimum value. So when (m+n)/2 is minimum, it implies (m+n) is minimum. Therefore, sum of numbers takes its minimum value when the numbers are equal.


When geometric mean is equal to arithmetic mean, it means the geometric mean is taking its maximum value. So when sqrt(m*n) is maximum, it means m*n is maximum. Therefore, product of numbers takes its maximum value when the numbers are equal.


Let’s see how to solve a difficult question using this concept.


Question: If x and y are positive, is x^2 + y^2 > 100?

Statement 1: 2xy < 100

Statement 2: (x + y)^2 > 200



We need to find whether x^2 + y^2 must be greater than 100.


Statement 1: 2xy < 100

Plug in some easy values to see that this is not sufficient alone.

If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100

If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100

So x^2 + y^2 may be less than or greater than 100.


Statement 2: (x + y)^2 > 200

There are two ways to deal with this statement. One is the algebra way which is easier to understand but far less intuitive. Another is using the concept we discussed above. Let’s look at both:


Algebra solution:


We know that (x – y)^2 >= 0 because a square is never negative.

So x^2 + y^2 – 2xy >= 0

x^2 + y^2 >= 2xy

This will be true for all values of x and y.


Now, statement 2 gives us x^2 + y^2 + 2xy > 200. The left hand side is greater than 200. If on the left we substitute 2xy with (x^2 + y^2), the left hand side will either become greater than or same as before. So in any case, the left hand side will remain greater than 200.

x^2 + y^2 + (x^2 + y^2) > 200

2(x^2 + y^2) > 200

x^2 + y^2 > 100


This statement alone is sufficient to say that x^2 + y^2 will be greater than 100. But, we agree that the first step where we start with (x – y)^2 is not intuitive. It may not hit you at all. Hence, here is another way to analyze this statement.


Logical solution:


Let’s try to find the minimum value of x^2 + y^2. It will take minimum value when x^2 = y^2 i.e. when  x = y (x and y are both positive)

We are given that (x+y)^2 > 200

(x+x)^2 > 200

x > sqrt(50)

So x^2 + y^2 will take a value greater than [sqrt(50)]^2 + [sqrt(50)]^2 = 100.

So in any case, x^2 + y^2 will be greater than 100. This statement alone is sufficient to answer the question.


Answer (B)


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