Inequality $\frac{x}{1+x} < \ln(1+x), \forall x>0$ 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.
 A: 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.
A: 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 $t=1$.
A: 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!
A: 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 $x=0$.
A: Starting with the (well-known?) inequality $1+u\le e^u$ for all $u$, let $u=-\log(x)$:
$$
\begin{align}
1+u&\le e^u\\
1-\log(x)&\le\frac1x\\
\frac{x}{1+x}&\le\log(x)
\end{align}
$$
A: 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 $x$.
