Question

If `x` and `y` are positive numbers, what is the minimum possible value of `(x+y)(1/x + 1/y)` ?

Original syntax:

If $x$ and $y$ are positive numbers, what is the minimum possible value of $(x+y)\left(\frac1x + \frac1y\right)?$

Answer 1

Four

Explanation:

`f(x,y) = x 1/x + x 1/y + y 1/x + y 1/y`

`frac{del f}{del x} = 1/y - y/x^2 = 0 Rightarrow x^2 = y^2`

`frac{del f}{del y} = -x/y^2 + 1/x = 0`

`f(x, x) = (x + x)(1/x + 1/x) = 2x * 2/x = 4`

`f(x, -x) = (x - x)(1/x - 1/x) = 0`

Now, let's begin again. Suppose with some loss of generality that `y = ax + b`.

`f(x, ax + b) = (x + ax + b)(1/x + 1/(ax + b)) = g(x)`

What are the "extremee"?

`g'(x) = (1 + a)(1/x + 1/(ax + b)) + (x + ax + b)(-1/x^2-frac{a}{(ax+b)^2}) = 0`

`(1 + a)(x(ax +b)^2 + x^2(ax + b)) + (x + ax + b)(-(ax+b)^2-ax^2) = 0`

#x(ax +b)^2 + x^2(ax + b) + ax(ax +b)^2 + ax^2(ax + b) -(x + ax + b)(ax+b)^2-ax^2(x + ax + b) = 0#

#(x + ax - x - ax - b)(ax +b)^2 + (x^2 +ax^2 - ax^2)(ax + b) -ax^3 = 0#

`- b(a^2x^2 + 2abx +b^2) + bx^2 = 0`

`b = 0 or (a^2 - 1)x^2 + 2abx +b^2 = 0`

`Delta = 4a^2b^2 - 4b^2(a^2 - 1) = 4b^2`

`x = frac{-a ± 1}{a^2 - 1} b`

`x_1 = frac{-b}{a + 1}` and `x_2 = frac{-b}{a-1}`

`g(x_1) = (frac{-b}{a + 1} + afrac{-b}{a + 1} + b)(1/frac{-b}{a + 1} + 1/(afrac{-b}{a + 1} + b)) = 0`

`g(x_2) = ( frac{-b}{a-1} + a frac{-b}{a-1} + b)(1/ frac{-b}{a-1} + 1/(a frac{-b}{a-1} + b))`

`g(x_2) = ( frac{-2b}{a-1} )(frac{a-1}{-b} + frac{a - 1}{-ab + ab - b})`

`g(x_2) = ( frac{2b}{a-1} )(frac{2(a - 1)}{b}) = 4`

Fortunately, all true paths lead to the same true consequence.

Answer 2

`4`

Explanation:

By symmetry `x=y=lambda` so substituting

`(2 lambda)(2/lambda) = 4`

Now supposing `x = lambda-delta` and `y = lambda+delta` we have

`(2lambda)(1/(lambda-delta)+1/(lambda+delta)) = (4 lambda^2)/(lambda^2-delta^2)`

As we can observe the minimum value for

`f(lambda) = (4 lambda^2)/(lambda^2-delta^2)`

is `4` for `delta = 0`

Answer 3

`4`

Explanation:

`(x+y)(1/x+1/y)=2+x/y+y/x`

From the arithmetic mean-geometric mean inequality, `(x/y+y/x)/2≥sqrt(x/y*y/x)`, or `x/y+y/x≥2`. Equality occurs when `x=y=1`.

Thus, `(x+y)(1/x+1/y)=2+x/y+y/x≥2+2=4`