I think you are saying that(1+h)^n ≥ h^n +1
because (1+h)^n will be at least h^n + xh^(n-1)+...+1
I believe h^n + 1 ≥ nh+1
all I would have to do is show x^y ≥ xy and the proof would be done. Do you think my method is sound?
Right: induction is great at an elementary level. A different (calculus-based) method shows the result to be true if ##h >-1## and ##n \geq 1## is real. However, I guess this would not be appropriate in a "pre-calculus" forum.The usual proof of this inequality is by induction on n. The proof by induction works even if ##h>-1\Rightarrow h+1>0##.(while your proof at post #3 works only if ##h>1##.)
This was the first proof by induction that I was taught , I remembered my high school years now.
if we constrain ##n\geq 1## to be rational, then the result follows from basic ##\text{GM}\leq \text{AM}## (proven by one of many elementary means). However to consider irrational values, some kind of analytic tools end up being needed, naturally.Right: induction is great at an elementary level. A different (calculus-based) method shows the result to be true if ##h >-1## and ##n \geq 1## is real. However, I guess this would not be appropriate in a "pre-calculus" forum.
I am interested in your calculus based method that proves it for ##n## real (and ##h>-1##). Can you give any hint?Right: induction is great at an elementary level. A different (calculus-based) method shows the result to be true if ##h >-1## and ##n \geq 1## is real. However, I guess this would not be appropriate in a "pre-calculus" forum.
I think the logarithm plus power series could work, but there has to be a more elegant solution with the mean value theorem or so.I am interested in your calculus based method that proves it for ##n## real (and ##h>-1##). Can you give any hint?
This proof has been brought to you by Hölder & Young....
##(1 +ph)^\frac{1}{p}##
## = (1 +ph)^\frac{1}{p}\cdot 1##
## = (1 +ph)^\frac{1}{p}\cdot 1^\frac{p-1}{p}##
## \leq \frac{1}{p} (1 +ph) + \frac{p-1}{p}(1)##
## = 1 +h##
by ##\text{GM} \leq \text{AM}##
(you just need analytic tools to prove the strong form of this geometric vs arithmetic mean inequality)
it is true that I used ##p## in a nod to Hölder's Inequality and these are Hölder conjugates in the exponents. That said, one doesn't actually need to know Hölder's Inequality (or Young) to get this -- I think Cauchy would have gotten it by just playing around with ##\text{GM} \leq \text{AM}##.
My personal preference would be to note that for any real ##n > 1## the function ##f(h) = (1+h)^n## is strictly convex on the set ##(-1,\infty)##, so ##f(h) > f(0) + f'(0) h ## for any ##h \neq 0, h > -1.## This requires a little bit of fairly elementary material on convexity; in this case we need to know that for a convex function ##f(h),## the tangent line to the graph of ## y = f(h)## lies on or below the graph at all legitimate ##h \neq 0##, and lies strictly below if ##f## is strictly convex.I'll take that at face value and show my preferred approach for the general case (a lot nicer than wikipedia in my view):
for real ##p \geq 1## and ##h \in(-1,\infty)## we have
##1 +ph \leq (1 +h)^p##
the right side is always positive so we only need to put effort into proving the case where the left hand side is positive.
in which case we can take pth roots and seek to prove the equivalent ##(1 +ph)^\frac{1}{p} \leq 1 +h##.
So we have
##(1 +ph)^\frac{1}{p}##
## = (1 +ph)^\frac{1}{p}\cdot 1##
## = (1 +ph)^\frac{1}{p}\cdot 1^\frac{p-1}{p}##
## \leq \frac{1}{p} (1 +ph) + \frac{p-1}{p}(1)##
## = 1 +h##
by ##\text{GM} \leq \text{AM}##
(you just need analytic tools to prove the strong form of this geometric vs arithmetic mean inequality)