Prove $\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8}$ As the title says, prove $$\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8},$$ for $n>1$. This inequality is from Erdős, "Problems and results on the theory of interpolation". I, Lemma 3.
My attempt: since $H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k} > \int_{1}^n\frac{1}{t}dt = \log(n),$ we then have
$$
\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} = \sum_{k=1}^{n-1}\frac{n^2 - 2kn + k^2}{2k} = \frac{n^2}{2}H_{n-1} - n(n-1) + \frac{n(n-1)}{4} > \frac{n^2\log(n)}{2} - \frac{3n(n-1)}{4}.
$$
Am I missing something here?

Your bound is much tighter for sufficiently large $n$.  Define $$f(n) = \frac{n^2 \log n}{8}, \\ g(n) = \frac{n^2 \log n}{2} - \frac{3n(n-1)}{4}.$$  Then the ratio $$\frac{g(n)}{f(n)} = 4 - \frac{6(n-1)}{n \log n}.$$  For $n \ge 5$, $\log n > 1.6$, hence $\frac{6(n-1)}{n \log n} < 3$, thus $g > f$.
In the case where $n \le 4$, we would directly compare the bound $\frac{n^2 \log n}{8}$ with the sum.
