How to prove $\ln(1+x)\ge x/(1+x/2)$ when $x > 0$?

Question

I think the equation establishes only when $x = 0$, so the condition should be x>=0. The first derivative of $\ln(1+x)-x/(1+x/2)$ is $0$ at $x=0$. So the inequality looks good to establish. Is there a better way to prove it?

Answer 1

Consider $$ f(x)=\ln(1+x)-\frac{2x}{2+x}=\ln(1+x)-2+\frac{4}{2+x} $$ Then $$ f'(x)=\frac{1}{1+x}-\frac{4}{(2+x)^2}= \frac{x^2}{(1+x)(2+x)^2} $$ Hence $f'(x)>0$ for $x>0$. Since $f(0)=0$, we are done.

Answer 2

Apply the mean value theorem on $f\colon[0,x] \to \mathbf R$ with $$f(t):= \ln(1+t)- \frac{t}{1+\frac{t}{2}}$$ where we assume $x>0$. Therefore whe have that there is a $\xi \in (0,x)$ with $$\ln(1+x)- \frac{x}{1+\frac{x}{2}}=f(x)-f(0)= f'(\xi)(x-0)=\frac{\xi^2}{(\xi +1)(\xi+2)}.$$ Since we have $\xi >0$ and $x>0$ we get the desired inequality $$\ln(1+x)- \frac{x}{1+\frac{x}{2}}>0.$$ For $x=0$ we have $\ln1 = \frac{0}{1+\frac{0}{2}}$ so the inequality $$\ln(1+x) \geq \frac{x}{1+\frac{x}{2}}$$ holds for alle $x\geq 0$.

Answer 3

To show $\ln(1+x) \ge x/(1+x/2) $.

Since $\ln(1+x) =\int_0^x \dfrac{dt}{1+t} $, this is equivalent to $\dfrac1{x}\int_0^x \dfrac{dt}{1+t} \ge \dfrac1{1+x/2} $.

Since $\dfrac1{1+x/2}$ is the value of $\dfrac1{1+t} $ at the midpoint, I'll combine the two halves.

$\begin{array}\\ \int_0^x \frac{dt}{1+t} &=\int_0^{x/2} \frac{dt}{1+t}+\int_{x/2}^x \frac{dt}{1+t}\\ &=\int_0^{x/2} \frac{dt}{1+t}+\int_0^{x/2} \frac{dt}{1+x-t}\\ &=\int_0^{x/2} dt\left(\frac{1}{1+t}+\frac{1}{1+x-t}\right)\\ &=\int_0^{x/2} dt\left(\frac{1+x-t+1+t}{(1+t)(1+x-t)}\right)\\ &=\int_0^{x/2} dt\left(\frac{2+x}{1+t+x-t+t(x-t)}\right)\\ &=(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+t(x-t)}\right)\\ &\ge(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+x^2/4}\right) \qquad\text{since }t(x-t) \le x^2/4\\ &=\dfrac{2+x}{(1+x/2)^2}\int_0^{x/2} dt\\ &=\dfrac{x}{2}\dfrac{4(2+x)}{(2+x)^2}\\ &=\dfrac{2x}{x+2}\\ &=\dfrac{x}{1+x/2}\\ \end{array} $

so that $\dfrac1{x}\int_0^x \frac{dt}{1+t} \ge \dfrac{1}{1+x/2} $.

Note that the only inequality used is that $t(x-t) \le x^2/4$ for $0 \le t \le x/2$.

If we use $t(x-t) \ge 0$, we can get an inequality going the other way:

$\begin{array}\\ \int_0^x \frac{dt}{1+t} &=(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+t(x-t)}\right)\\ &\le(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x}\right)\\ &=\dfrac{2+x}{1+x}\int_0^{x/2} dt\\ &=\dfrac{x(2+x)}{2(1+x)}\\ \end{array} $

so that $\ln(1+x) \le \dfrac{x(2+x)}{2(1+x)} $.

The difference of the two bounds is

$\begin{array}\\ \dfrac{x(2+x)}{2(1+x)}-\dfrac{2x}{x+2} &=\dfrac{x(2+x)(x+2)-2x(2(1+x))}{2(1+x)(x+2)}\\ &=\dfrac{x(x^2+4x+4-(4+4x))}{2(1+x)(x+2)}\\ &=\dfrac{x^3}{2(1+x)(x+2)}\\ &\le \dfrac{x^3}{4}\\ \end{array} $

Answer 4

Bounds using the Taylor series can be used but it is far more work. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

$\exp(x) \le \sum_{k=0}^{n-1} \lfrac1{k!} x^k + \lfrac2{n!} x^n$ for every real $x \in [0,\lfrac{n+1}{2}]$.

In particular the one we need is:

$\exp(x) \le 1+x+\lfrac12x^2+\lfrac16x^3+\lfrac1{12}x^4$ for every real $x \in [0,2.5]$.

Just comparing derivatives only works for $x \in [0,\ln(2)]$. But using a geometric series to bound the error gives $\sum_{k=n}^\infty \lfrac1{k!} x^k \le \lfrac1{n!}x^n \sum_{k=0}^\infty (n+1)^{-k} x^k \le \lfrac1{n!}x^n \sum_{k=0}^\infty 2^{-k} = \lfrac2{n!}x^n$ as desired.

Now take any real $y \ge 0$. Let $x = \lfrac{2y}{2+y} \in [0,2)$. Then $y = \lfrac{2x}{2-x}$ and all we need to prove is

$1+x+\lfrac12x^2+\lfrac16x^3+\lfrac1{12}x^4 \le 1+\lfrac{2x}{2-x}$.

This is equivalent to:

$(1+x+\lfrac12x^2+\lfrac16x^3+\lfrac1{12}x^4)(2-x) \le 2+x$.

Which simplifies to an obvious inequality.

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Notes

The desired result is actually a Padé approximant for $\ln$, and egreg has already shown how to prove it directly. Note that it can be used to prove the Padé approximant for $\exp$, which is especially useful when we want an upper bound that the Taylor series does not give. Specifically:

$\exp(x) \le \lfrac{2+x}{2-x}$ for every real $x \in [0,2)$.

Which in fact clearly follows from the above inequalities as well. So we have two methods to prove this upper bound for $\exp$, and it is a matter of taste which is easier.

Answer 5

Take the exponential of both terms and re arrange the equation in order to be written as

$$1 + x \geq \large e^{\frac{2x}{2+x}}$$

Now you can use the Taylor series for the exponential, stopping at first order, to get

$$1 + x \geq 1 + \frac{2x}{2+x}$$

$$x \geq \frac{2x}{2+x}$$

$$1 \geq \frac{2}{2+x}$$

$$2 + x \geq 2$$

Which holds $\forall x \geq 0$