Question 41

# Let x and y be two positive numbers such that $$x + y = 1.$$Then the minimum value of $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ is

Solution

Approach 1:

The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.

=>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ =12.5

Approach 2:

$$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$(x+1/x+y+1/y)^2$$ - $$2*(x+1/x)(y+1/y)$$

Let x+1/x and y+1/y be two terms. Thus (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $$\sqrt{(x+1/x)(y+1/y)}$$ would be their Geometric Mean (GM).

Therefore, we can express the above equation as $$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$4AM^2$$ - $$GM^2$$. As AM >= GM, the minimum value of expression would be attained when AM = GM.

When AM = GM, both terms are equal. That is x+1/x = y +1/y.

Substituting y=1-x we get

x+1/x = (1-x) + 1/(1-x)

On solving we get 2x-1 = (2x-1)/ x(1-x)

So either 2x-1 = 0 or x(1-x) = 1

x(1-x) = x * y

As x and y are positive numbers whose sum = 1, 0<= x, y <=1. Hence, their product cannot be 1.

Thus, 2x-1 = 0 or x=1/2

=> y = 1/2

So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ = 12.5