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All similar proofs I could find show that $(1+1/n)^n$ is increasing for positive integers values of $n$ only, or show that the derivative of $(1+1/x)^x$ is positive for all $x \in \mathbb{R}_{>0}$.

I'm looking for a proof that $x<y \implies (1+1/x)^x < (1+1/y)^y$ for all $x,y \in \mathbb{R}_{>0}$ that doesn't use the derivative of $a^x$ nor the derivative of $\log_a (x)$

My attempt so far:

Let $x,y \in \mathbb{R}_{>0}$ such that $x<y$. Let

$$ \begin{align} J&= \frac {\left(1+\frac{1}{y}\right)^y}{\left( 1+ \frac{1}{x} \right)^x}\\ &= \frac {(\frac{y+1}{y})^x(1+\frac{1}{y})^{y-x}}{(\frac{x+1}{x} )^x}\\ &= {(\frac{xy+x}{xy+y})^x(1+\frac{1}{y})^{y-x}} \\ &= {\left(1 - \frac{y-x}{xy+y}\right)^x\left(1+\frac{1}{y}\right)^{y-x}} \end{align} $$

Showing that $\left(1 - \frac{y-x}{xy+y}\right)^x > \frac{1}{\left(1+\frac{1}{y}\right)^{y-x}}$ is what I'm missing

Benjamin Wang
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Hussein Aiman
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3 Answers3

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Well, since you know Bernoulli's inequality, then for $0<x<y$, note: $$\left(1+\frac1y \right)^{\frac{y}x}>1+\frac{y}x\cdot\frac1y=1+\frac1x$$

Macavity
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  • That just hides the question for how to prove Bernoulli for real exponents. There are easy proofs of Bernoulli for the exponents which are positive integers (induction.) – Thomas Andrews Mar 01 '24 at 15:17
  • Basically the binomial theorem is all you need for integer exponents, but the question becomes harder for real exponents because, even when the infinite binomial converges, it has negative terms. – Thomas Andrews Mar 01 '24 at 15:19
  • @ThomasAndrews At the core of the issue, irrational exponentiation definition itself will require tools beyond "elementary", like limits and continuity, and then those tools are enough to extend Bernoulli's inequality (using weighted AM-GM, or concavity of log, say) from rational exponents to all real ones. – Macavity Mar 01 '24 at 16:10
  • You might just prove $(1+x)^p\geq (1+xp/q)^q$ for $p,q$ positive integers, $p\geq q,$ and then use continuity of $(1+x)^r$ in $r$ for irrational $r.$ You don't need logarithms, though they definitely make things easier. – Thomas Andrews Mar 01 '24 at 17:39
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I managed to prove the part I'm missing if I add the further assumption that $x < y < 2x$.

$y - x < x$ and $\frac{x}{y - x} > 1$

Let $K = \frac{y-x}{xy+y} = \frac{y-x}{(x+1)y}$ and $\alpha=y-x$

$K = \frac{\alpha}{(x+1)(x+\alpha)} = \frac{\alpha}{x^2+\alpha x+ x + \alpha}$

$\frac{1}{K} = \frac{x^2+\alpha x+ x + \alpha}{\alpha} = \frac{x^2+\alpha x+ x}{\alpha} + 1 >1$ and $0 < K < 1$

Since $-1 < -\frac{y-x}{xy+y} < 0$ and $\frac{x}{y-x} > 1$, we can use Bernoulli's inequality.

$(1-\frac{y-x}{xy+y})^{\frac{x}{y-x}} > 1 - (\frac{y-x}{xy+y})(\frac{x}{y-x}) = 1 - \frac{1}{y+1} = \frac{y}{y+1}$

$(1-\frac{y-x}{xy+y})^x = \Big((1-\frac{y-x}{xy+y})^{\frac{x}{y-x}} \Big)^{y-x} > (\frac{y}{y+1})^{y-x} = \frac{1}{(1+\frac{1}{y})^{y-x}}$

Therfore, $\frac {(1+\frac{1}{y})^y}{( 1+ \frac{1}{x} )^x} = {(1 - \frac{y-x}{xy+y})^x(1+\frac{1}{y})^{y-x}} > 1$ and $(1+\frac{1}{y})^y > ( 1+ \frac{1}{x} )^x$

Hussein Aiman
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$$f(x)=x\log(\frac{x+1}{x})\Rightarrow f'(x)=\log(\frac{x+1}{x})-\frac{1}{1+x}\geq 0$$ from $\log y\leq y-1$ applied to $y=\frac{x}{x+1}.$

Letac Gérard
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