A pair of inequalities from analysis

  • #1
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Homework Statement



Prove ln(1+x)>=(x)/Sqrt(1+x)

Prove (x-1)^2>=x((ln(x))^2)
For x>0

Homework Equations





The Attempt at a Solution



I have tried using MVT, but i only end up with more inequalities that i cannot seem to prove... Another idea that works but i cannot prove exactly why it works is saying that since both sides "start" at the same value and the value of one derivative is always greater than the other implies that the LHS is always greater than the RHS. But aside from intuition, i have no proof as to why this is true! Any help would be appreciated!
 

Answers and Replies

  • #2
85
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i dunno how to solve it analytically but can you just say ln(1+x)< 1+x > x > x/sqrt(x+1). i reckon no.
 
  • #3
102
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hmm no that wont do. for the first, is it enough to say that since at x=0 both sides are equal and the derivative of one is always greater than the other? A similar reasoning might work for x>1 by taking the first 3 derivatives...
 
  • #4
Gib Z
Homework Helper
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For the first one, I would try a linear approximation of ln(1+x) at 0.
 
  • #5
VietDao29
Homework Helper
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I have tried using MVT, but i only end up with more inequalities that i cannot seem to prove... Another idea that works but i cannot prove exactly why it works is saying that since both sides "start" at the same value and the value of one derivative is always greater than the other implies that the LHS is always greater than the RHS. But aside from intuition, i have no proof as to why this is true! Any help would be appreciated!


Ok, the proof for that goes like this:
Say f(x) and g(x) are the 2 functions such that:

f(x0) = g(x0)

And f'(x) >= g'(x), for all x >= x0

Let h(x) = f(x) - g(x), now, we have:

h(x0) = f(x0) - g(x0) = 0
h'(x) = f'(x) - g'(x) >= 0 for x >= x0

Since it's derivative is non-negative for x >= x0, so on that interval h(x) is an increasing function.

So for x >= x0, we have:

h(x) >= h(x0) = 0

~~> f(x) - g(x) >= 0 ~~> f(x) >= g(x), for x >= x0

The idea above can be used to prove the first one.

Are you sure about the second problem? It does not seem true at all. :bugeye:
 
  • #6
102
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Thats sweet! Just what i needed. Yup, the second is true all right. The derivative method works fine by taking the first three derivatives, but its a little messy and only works on (1, infinity). I managed to find something on (0,1] but its also messy. Any elegant ideas?
 

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