Proving question, anyone?

Question

Given x + y =1. Prove that (1/x) + (1/y) ≥ 4
Thank you for your help!

Answer 1

Let's use calculus. x+y=1, so y=1-x. Replace y with this in the function g(x,y) = 1/x + 1/y to get a function f(x):

f(x)=1/x + 1/(1-x) = 1/(x*(1-x)).

We care about values of x between 0 and 1, since otherwise, as noted by other people, the result just isn't true. As x goes to 0 or 1 (inside the interval (0,1)), f(x) goes to infinity. So let's find the minimum.

f'(x) = (2x-1) / (x^2*(x-1)^2),

which is zero at x=1/2. We know this has to be where the minimum of f(x) occurs, but you can check it with a sign chart (first derivative test) or by the second derivative test. Hence, the minimum value of f(x) is f(1/2)=4, i.e.

f(x) = 1/x + 1/(1-x) = 1/x + 1/y >= 4 for x,y>0 with x+y=1.

Answer 2

Given x + y = 1. Prove that (1/x) + (1/y) >= 4.

It's false. Let x = -5 and y = 6.

Then (-1/5) + (1/6) = -6/30 + 5/30 = -1/30, and -1/30 is not greater than or equal to 4. Perhaps you forgot to put an added condition?

*******
I'm going to take a wild guess and you forgot to add that x and y are greater than 0.

Proof:

The range of the function f(z) = z^2 would be all values greater than or equal to z; therefore

z^2 >= 0.

This is true no matter what z is. Let's let z = (x - [1/2]). Then

(x - [1/2])^2 >= 0. Expanding the left hand side,

x^2 - x + 1/4 >= 0

Moving the x^2 and the -x to the right hand side,

1/4 >= x - x^2

Factoring x on the right hand side,

1/4 >= x(1 - x)

And now, noting that since x + y = 1, y = 1 - x, so replacing y with (1 - x) yields

1/4 >= xy

Multiplying both sides by 4,

1 >= 4xy

Replacing 1 with (x + y) [Since it is given that x + y = 1]

x + y >= 4xy

At this point, we multiply both sides by (1/(xy)); this is perfectly safe to do, because x > 0 and y > 0, so we don't have to worry about things like xy being 0, or any sign changes after multiplying the inequality.

(1/[xy]) (x + y) >= 4

x/[xy] + y/[xy] >= 4
(1/y) + (1/x) >= 4, which is the same as

(1/x) + (1/y) >= 4.

Answer 3

i think the question is:
for x , y > 0 , if (x + y) = 1 ,prove that 1/x + 1/y ≥ 4
okay,we have this well-known inequality for x,y ≥ 0:

(x + y)/2 ≥ 2 / (1/x + 1/y)

just put x + y = 1 then:
1/2 ≥ 2 / (1/x + 1/y)
=> (1/x + 1/y) ≥ 1/2 * 1/2
=> 1/x + 1/y ≥ 4
and equality happens when x = y , so it happens when x = y = 1/2

good luck

Answer 4

I think puggy is making it too dificult. Here is my take:

Let x=y=0.5 for the moment:

so 1/x = 2 & 1/y = 2.

Let x be any number smaller than or equal to 0.5, y be any number greater than of equal to 0.5, keep in mind that x + y = 1

1/x >= 2 ----------(1)
1/y <= 2 ----------(2) multiply both sides with -1
-1/y >= -2 --------(3)

Here you can then perform the operation on equations (1) - (3)

1/x + 1/y >= 4........Q.E.D.


XR

Answer 5

im not sure what u mean by that question. where did u get it from? and are u sure u copied the question out correctly?

anyway... did u try factorising the fractions?
you get:
(y + x) / (xy) ≥ 4
(x + Y) ≥ 4xy
1 ≥ 4xy
which is only true when.... etc.... urrgh im lost.

Answer 6

I dont know