Proving f(x)=1/(1+x) is Greater than 1-x for x>=0

[sean/mac]

Let f(x)=1/(1+x)

Use the Mean Value Theorom (for the derivative of a function) to prove that f(x)>=1-x for x>=0

also

Mean Value Theorom states:

[f(b)-f(a)]/ [b-a]= f'(c) where c is an element of [a,b]

 

[daniel_i_l]

what do you get for f(x) when you use the MVT with b=x, a=0? What are the limits on c?

 

[sean/mac]

[f(b) - f(a)]/[b-a]=f'(c) for b=x and a=0

gives

[1/(1+x) - 1]/x=-1/(1+c)^2

which when i do the algebraic manipulation gives

1+x=(1+c)^2

i don't know how to make a relation between 1-x and 1/(1+x)

 

[HallsofIvy]

[f(b) - f(a)]/[b-a]=f'(c) for b=x and a=0

gives

[1/(1+x) - 1]/x=-1/(1+c)^2

which when i do the algebraic manipulation gives

1+x=(1+c)^2

i don't know how to make a relation between 1-x and 1/(1+x)

What is the largest possible value for 1/(1+c)²?

 

[DavidWhitbeck]

[f(b) - f(a)]/[b-a]=f'(c) for b=x and a=0

gives

[1/(1+x) - 1]/x=-1/(1+c)^2

You already went too far, you want to keep f (don't substitute) on the lhs to get that inequality, so keep it like this:
(f(x) -1)/x = -1/(1+c^2)
and then follow HallsofIvy's hint.