Prove that $\ln x \leq x - 1$ I need help with this proof for my real analysis class. it is on the practice sheets and we do NOT get an answer. I proved $\ln(x) < x−1$ for all $x>1$ by contradiction but cannot do this one. 
Prove that $\ln(x) \leq x−1$ for all $x>0$.
i believe you need to use MVT, I cannot use the famous inequality $e^x>x+1$ for all $x>0$.
 A: Apply what you've already proven using $\ln(\frac{1}{x})=-\ln(x)$.
A: To prove: $lnx \leq x - 1$ for $x > 0$. Since you've done the part for $x > 1$, we show: the inequality true for $x \in (0, 1]$. At $x = 1$ we have equality, so consider $x \in (0, 1)$. Then $0 < 1 - x < 1$. So, using a power series expansion for $ln(1 - x)$ at $1 - x$ we have:
$\ln x = \ln(1 - (1-x)) = -(1-x) - \dfrac{(1-x)^2}{2} - \dfrac{(1-x)^3}{3} - \dfrac{(1-x)^4}{4} -\cdots-\dfrac{(1-x)^n}{n}-\cdots < -(1-x) = x - 1$
A: If $x=1$ it's true. Suppose $0<x<1$. Define $f(x)=\ln x -x +1$. Then $f'(x)=\frac{1}{x}-1$. As $0<x<1$, $f'(x)>0$, then $f$ is crescent. So $f(x)<f(1)$ for $0<x<1$.As $f(1)=0$, $\ln x -x +1<0$, therefore $\ln x ≤ x - 1$ for $0<x≤1$. 
A: Note that, for $y>1$  $$\int_1^y\frac{dt}{t^2}\leqslant\int_1^y\frac{dt}t\leqslant\int_1^y dt$$  gives $$\tag 1 1-\frac 1 y \leqslant \log y \leqslant y-1$$
with equality iff $y=1$. If $0<y'<1$, write $y= \frac 1{ y'}$ and use the above to get $$1 - y'\leqslant \log \frac{1}{{y'}}\leqslant \frac{1}{{y'}} - 1$$ and multiply by $-1$ to get $(1)$ is true also in this case.
