How might one prove the following inequality?

Solution 1:

Let $f(x)=x^{r+1}+\log(r+1)-\sum_{i=1}^r{x^i/i}$. We need to show that $\min_{x>0}f(x)>0$.

Differentiate and set to $0$: $$f'(x)=(r+1)x^r-\sum_{i=0}^{r-1}{x^i}=0$$ Clearly $f'(t)>0$ for $t\geq 1$, so solution(s) for $f'(x)=0$ lies in $(0,1)$ (at least one exists since $f'(0)<0$ and $f'$is continuous). Use $x^r=\sum_{i=0}^{r-1}{x^i}/(r+1)$, $x<1$ and telescoping product for $\log$:

$$f(x)=x^{r+1}+\log(r+1)-\sum_{i=1}^r{x^i/i}=x\cdot(\sum_{i=0}^{r-1}{x^i}/(r+1))+\log(r+1)-\sum_{i=1}^r{x^i/i}=\sum_{i=1}^{r}x^i(1/(r+1)-1/i)+\log(\prod_{i=1}^r\frac{1+i}{i})>\sum_{i=1}^{r}(1/(r+1)-1/i)+\sum_{i=1}^r\log(1+1/i)=\frac{r}{r+1}-\sum_{i=1}^r\big(1/i-\log(1+1/i)\big)>\frac{r}{r+1}-\gamma>0$$ for $r\geq 2$ (where $\gamma\approx0.577$ is Euler–Mascheroni constant). Separately check the case $r=1$ to get $f(x)\geq \log(2)-1/4>0$.