Prove $\sum\limits^{n-1}_{i=1} i\ln (i) \leq \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}$ for $n\geq 1$. I'm quite stuck, truly. I have tried proving it directly, using induction, showing that $$n\ln (n) \leq \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big)$$
as well as using the fact that $$\sum ^n_{i=1}i\ln (i)\leq \sum ^n_{i=1}i\ln (n)=\frac{n(n-1)}{2}\ln (n
)$$
but haven't been able to achieve much. The inequality does hold though, I've checked using desmos.
 A: Look at the function $f(x)=x\ln{x}$ and recall Riemann sums (i.e. lower Riemann sum or lower Darboux sum). Given $f'(x)=\ln{x}+1$, $f(x)$ is monotone ascending for $x\geq1$. As a result
$$\sum^{n-1}_{i=1} i\ln{i}=
\sum^{n-1}_{i=1} i\ln{i} \left(i+1-i\right)=\\
\sum^{n-1}_{i=1} f(i) \left(i+1-i\right)\leq \int_{1}^{n}f(x)dx=\\
\frac{x^2}{4}(2\ln{x}-1)\Big|_{1}^n = 
\frac{n^2}{4}(2\ln{n}-1)+\frac{1}{4}=\\
\frac{n^2\ln{n}}{2}-\frac{n^2}{4}+\frac{1}{4}$$
A: It's not clear to me what the sticking point is.  If you have proven that $$n\ln (n) \leq \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big)$$ then you are basically done, as this shows that the change in the left side of your desired inequality is less than the change of the right side.  
To be explicit, let $f(n):=\sum_{i=1}^{n-1} i \ln (i)$ and let $g(n):= \frac{n^2}{2}\ln (n)-\frac{n^2}{4}-\frac{1}{4}$.  Then $f(n+1)-f(n)$ is the left side of the above inequality, and $g(n+1)-g(n)$ is the right side.  By induction, if $f(n)\leq g(n)$, then $$f(n+1) = \left(f(n+1)-f(n)\right)+f(n) \leq \left(g(n+1)-g(n)\right) + g(n) = g(n+1)$$ and together with $f(2)=0\leq 2\ln(2)-\frac{5}{4} = g(2)$ you're done.  (The reason I chose $n=2$ as a base case rather than $n=1$ is because it makes the following argument work more smoothly.  The $n=1$ case is trivial.)
The hard part is showing that the top inequality holds.  Here's one way to do so:
$$
\begin{align}
& \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big) \\
= & \frac{n^2}{2}\left(\ln (n+1)-\ln(n)\right) + n\ln(n+1) + \frac{1}{2}\ln(n+1) - \frac{n}{2} - \frac{1}{4} \\
= & \frac{n^2}{2}\ln (1+\frac{1}{n}) + n\ln(n+1) + \frac{1}{2}\ln(n+1) - \frac{n}{2} - \frac{1}{4}
\end{align}
$$
Using $\ln(1+\frac{1}{n}) \geq \frac{1}{n}-\frac{1}{2n^2}$ (from the Taylor series expansion of $\ln$) on the first term, $\ln(n+1)\geq\ln(n)$ on the second term, and $\ln(n+1)\geq 1$ (for $n\geq 2$) for the third term, we find that the above is 
$$\geq\frac{n}{2}-\frac{1}{4} + n\ln(n)+\frac{1}{2}-\frac{n}{2}-\frac{1}{4} = n\ln(n)$$as desired. 
