4

Prove that $\frac{x}{1+x} < \ln(1+x), \forall x>0$. I wrote it as $e^x < 1 + x + (1+x)^x$ to see if it would make it any simpler. I do not think induction would work since that only works for natural numbers (right?). I also tried writing it as $1 + x + (1+x)^x - e^x > 0$, showing that for a small $x$, the expression would always be positive, and since $x, e^x$, and $(1+x)^x$ are strictly increasing for $x>0$, the expression would always stay positive. But I do not know what $x$ to use for this situation. Thank you for your help.

jimjim
  • 9,675
AnthonyT
  • 41

6 Answers6

5

Multiply through by $1+x$ and differentiate both sides. On the left, you get $1,$ on the right you get $1 + \log (1+x)>1.$ Since the two sides are equal at $0,$ you are done.

Igor Rivin
  • 25,994
  • 1
  • 19
  • 40
  • So if the the inequality holds true for their derivatives, it must hold true for them, as well? Also, would taking the limit as x approaches 0 of the expression work, too? – AnthonyT Dec 03 '13 at 22:09
  • 4
    If two functions are equal at a point, and the derivative of one is greater than the derivative of the other, the first function is greater than the second to the right of the point... Limit is not enough, you need the inequality everywhere. – Igor Rivin Dec 03 '13 at 22:11
  • Ah, I see then. Thank you for that. – AnthonyT Dec 03 '13 at 22:12
  • @AnthonyT : Those are important points mentioned by Igor, without them it is false http://math.stackexchange.com/questions/591636/is-fxgx-iff-frac-operatorname-d-operatorname-d-x-fx-frac-opera/591642?noredirect=1#591642 – jimjim Dec 03 '13 at 22:32
  • +1, made me realise something! – jimjim Dec 03 '13 at 22:33
  • Good counterexample provided, thank you for asking that. – AnthonyT Dec 03 '13 at 22:35
5

Equivalently you want to prove that $\dfrac{t-1}{t}=1-\dfrac 1 t < \log t, \forall t>1$

But when $t\geqslant 1$ $$\log t =\int_1^t \frac{dy}y\geqslant \int_1^t \frac{dy}{y^2}=1-\frac 1 t$$ with equality only at $t=1$.

Pedro
  • 122,002
2

Since $(1+x) > 0$ ($x > 0$), you can multiply both sides by $(1+x)$: $$\frac{x}{1+x} < \log(1+x) \Rightarrow x < (1+x)\log(1+x)$$ After, you can apply exponential function to both sides: $$x < (1+x)\log(1+x) \Rightarrow e^x < e^{1+x}(1+x) \Rightarrow e^x < ee^x(1+x)$$ You can simplify $e^x$ from both sides, since $e^x > 0$, and then you get that: $$1 < e(1+x)$$ Then:

$$ex + e - 1 > 0 \Rightarrow x + 1 - \frac{1}{e} > 0 \Rightarrow x > \frac{1}{e} - 1$$

Since, $\frac{1}{e} - 1 < 0$ and $x > 0$, then you have the proof you want!

the_candyman
  • 14,064
  • 4
  • 35
  • 62
2

Let $f(x)=\frac{x}{1+x}-\ln(1+x)$. Then $f'(x)=-\frac{x}{(1+x)^2}$, hence $f$ is stricly decreasing. Now $f(0)=0$.

This shows that the inequality holds if $x>-1$, equality is achieved only in $x=0$.

Michael Hoppe
  • 18,103
  • 3
  • 32
  • 49
  • $f$ is only strictly decreasing for $x>0$. So the implication in your final sentence that the inequality fails for $x\in(-1,0)$ is wrong. – Marc van Leeuwen Dec 04 '13 at 13:24
2

Starting with the (well-known?) inequality $1+u\le e^u$ for all $u$, let $u=-\log(x)$: $$ \begin{align} 1+u&\le e^u\\ 1-\log(x)&\le\frac1x\\ \frac{x}{1+x}&\le\log(x) \end{align} $$

robjohn
  • 345,667
0

Given that $\mathbb{ln}(1+x) = \int_{1}^{1+x}\frac{1}{t}dt$ we know that $\mathbb{ln}(1+x)$ is just the area under the curve $\frac{1}{t}$ between $1$ and $1+x$.

Since $\frac{1}{t}$ is strictly decreasing we know that it reaches its minimum at $1+x$ on this interval. The interval also has length $(1+x)-1=x$ and so the area under the curve is strictly greater than the area of the rectangle having height $\frac{1}{1+x}$ and width $x$.

Patrick
  • 2,106