# If x+y=1 and both x and y are positive real numbers, prove that (x+(1/x))^2+(y+(1/y))^2 will always be AT LEAST 12.5.HINT: when x and y both equal 0.5, (x+1/x)^2+(y+1/y)^2=12.5MORE HINTS: you...

*If x+y=1 and both x and y are positive real numbers, prove that (x+(1/x))^2+(y+(1/y))^2 will always be AT LEAST 12.5.*

*HINT: when x and y both equal 0.5, (x+1/x)^2+(y+1/y)^2=12.5MORE HINTS: you must prove that when a+b= a fixed value, a^2+b^2 is smallest when a=b*

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Since x+y=1 then y=1-x so your equation is

`(x+1/x)^2+((1-x)+(1/(1-x)))^2 `

`= x^2 + 2(x)(1/x) + 1/x^2 + (1-x)^2 + 2(1-x)/(1-x) + 1/(1-x)^2`

`=x^2+2+1/x^2 + 1 - 2x + x^2 + 2 + 1/(1-x)^2`

=2x^2-2x+5+1/x^2+1/(1-x)^2

No for x and y to be positive and x+y=1 then 0<x<1 and 0<y<1

Take the derivative.

`f'(x)=4x-2-2/x^3+2/(1-x)^3` We are looking for the minimum.

f'(x)=0 when `4x-2-2/x^3-2/(1-x)^3=0`

`4x-2=2/x^3-2/(1-x)^3`

`(4x-2)(x^3)(1-x)^3=2(1-x)^3-2x^3`

`(4x-2)x^3(1-x)^3-2(1-x)^3+2x^3=0`

x=1/2 is the only real root of this equation

`f''(x)=4+6/x^4-6/(1-x)^4`

`f''(1/2)=4+6(2)^4-6(2)^4=4 gt0` so this is a minimum.

f(0) is undefined, and f(1) is undefined. `lim_(x->0)f(x)=lim_(x->1)f(x)=+oo` .

`f(1/2)=(1/2+2)^2+(1/2+2)^2=6.25+5.25=12.5`

So `f(x)>=12.5`