Show that $\frac1x+\log x-\log(x+1)\ge0$ for $x\ge1$ without differential calculus I can't seem to solve this exercise. I want to show that: $\frac1x+\log x-\log(x+1)\ge0$ for $x\ge1$.
I've tried looking for bounds on $\log x-\log(x+1)$, but there's nothing promising I can derive. I'm thinking of multiplying out $x$, yielding:
$$1+x\log x-x\log(x+1)\ge0$$
I believe, I can bound $x\log(x)$ by $ex-e$, but it doesn't seem sufficient, because I cannot find a good bound on $x\log(x+1)$.
I can solve it with differential calculus (indeed, I can check it for $x=0$ then differentiate), but I'd like to see if there is a way to solve it without. We've not been introduced to the formal definitions of differentiation etc. yet.

$$\frac1x+\log x-\log(x+1)\ge0$$
$$\iff\frac1x\ge-\log x+\log(x+1)=\log\frac{x+1}x=\log\left(1+\frac1x\right)$$
Define $y=\frac1x$, so that $0<y\le1$:
$$\iff y\ge\log(1+y)$$
Now define $z=y+1$ so that $y=z-1$ and $1<z\le2$:
$$\iff z-1\ge\log z$$
which is an inequality you have.
