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 $(x+y)\left(\frac1x + \frac1y\right)?$

3 Answers
May 10, 2017

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.

May 10, 2017

#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#

May 13, 2017

#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#