Application of Lagrange Theorem

Yep , that's ok , infact a fine proof for me at least

I meant to say some x₁ , x₂ in the interval (0,p) and (p,1) respectively does the job

where f(p) = 0.5

Me dear friend , if " f ( 12 ) ≠12 " , how is equality condition achieved ?

@ Ricky Nice work

I had the following

let f(1/2) = K

so Applying LMVT both sides we get

2K = f'(x₁) and 2(1-k) = f'(x₂)

so 12K = 1f'(x₁)

12(1-k) = 1f'(x₂)

so 1f'(x₁) + 1f'(x₂) ≥ 2 (by AM≥HM)

If f(1/2) ≠1/2 , then also some other x₁ , x₂ satisfy the equality condition ...

My head obviously agrees with you Sir " :) " , but my heart most certainly doesn't " :( ". Please forgive me for utter nonsense , Sir . . Still , I think my proof's a little bit more " fullproof " , Sir , please respond if you enjoy it as well :) :)

Let another function " g " be defined such that ,

g ( x ) = 1 - f ( x )

It is now clear as water that ,

g ( 0 ) = 1 ............ And g ( 1 ) = 0

So , " g " must have a fixed point lying between ( 0 , 1 ) .

Hence , let that fixed point be " c " .

So , g ( c ) = 1 - f ( c ) = c

Or , f ( c ) = 1 - c .

Apply Langrange's Theorem for two intervals , " ( 0 , c ) " and " ( c , 1 ) " .

Clearly , for some " x₁ , x₂ " in the respective said intervals ,

f ' ( x₁ ) = 1 - cc

And , f ' ( x₂ ) = c1 - c

Clearly , " f ' ( x₁ ) + f ' ( x₂ ) = c1 - c + 1 - cc ≥ 2"

And obviously , " 1f ' ( x₁ ) + 1f ' ( x₂ ) = 1 - cc + c1 - c ≥ 2 "

Both inequalities are readily established by " AM - GM " inequality , with equality being achieved at " c = 12 " .

I don't know how obvious it is , but spotting the equality condition , I immediately thought of using a point " x " for which " f ( x ) = 12 " in Langrange's theorem and solved the initial question .

First, Rohan's solution:

By Intermediate Value Theorem, there will exist an x_o, 0<x_0<1 such that f(x_0)= \frac{1}{2}

No by Lagrange's, we will find x_1 \in (0,x_0) such that f'(x₁) = 1/2x_o

and x_2 \in (x_0,1) with f'(x_2) = 1/2(1-x₀)

Its easy to see that 1/f'(x₁) + 1/f'(x₂) = 2.

Now f'(x₁) + f'(x₂) = 1/2 [ 1/x_o + 1/(1-x₀) ] ≥ 1/2 X 4 = 2 (by Cauchy Schwarz inequality)

both equalities occur if f(1/2) = 1/2

I am really sorry Sir , but it seems to me as if I don't understand a bit what you are trying to say sir . Can you please write down the full thing , sir ? Obviously I know that you have precious time , but still , if possible , plz do write the solution sir.

But sir , that pair only proves the equality , doesn't it ? Does it prove the inequality as well , sir ?

I said , there is at least one pair of values , not that any " x₁ , x₂" will satisfy the conditions .

Seems i am not getting ur question..

Actually f'(x) can be negtive , so the least value is not 2 , is it?

Okies , I think I have written some confusing and wrong statements . I 'll restate the problem .

Let " f " be a real valued , differentiable function such that ,

f ( 0 ) = 0 ; And f ( 1 ) = 1 .

Prove that there exists at least one pair of " x₁ , x₂ E ( 0 , 1 ) " such that , for the above said function , the following inequalities hold simultaneously .

1 . 1f ' ( x₁ ) + 1f ' ( x₂ ) ≥ 2

2 . f ' ( x₁ ) + f ' ( x₂ ) ≥ 2

Ever so sorry , both " f ( x ₁ ) " and " f ( x ₂ ) " have to be positive . Thank you .

counter example..

Ricky, i wouldnt bet on that

oh heck, pretty careless of me again. What you said about the fixed pointwould be true if we had for instance, f(0)=1 and f(1)=0. I got mixed up with this. I had the same soln as Rohan's.

let c be a point such that f(c) = c 0<c<1

so f(1) - f(c)/1-c = f'(x₁)

f(c) - f(0) / c = f'(x₂)

we prove both f'(x₁) + f'(x₂) = 2 and 1/f'(x1) + 1/f'(x2) = 2

@ sir is it right?