Find the largest $\lambda$ such that $$\sum_{k=1}^n x_k^3\left(x_{k}-x_{k-1}\right)\ge \frac{1}{4}+\frac{\lambda}{n}$$ holds for any $n \in \mathbb{N}$ and any $x_0,x_1,x_2,\cdots, x_n\in\mathbb{R}$ satisfying $$0=x_0\le x_1\le x_2\le \cdots\le x_n=1.$$
It's obvious that the problem is to estimate the Darboux sum of $f(x)=x^3.$
Note that for any $a,b$, it holds that $$a^3(a-b)-\frac{1}{4}(a^4-b^4)=\frac{1}{4}(a-b)^2((a+b)^2+2a^2)\ge0,$$namely $$a^3(a-b)\ge \frac{1}{4}(a^4-b^4).$$
Therefore $$\sum_{k=1}^n x_k^3\left(x_{k}-x_{k-1}\right)\ge \frac{1}{4}\sum_{k=1}^n \left(x_{k}^4-x_{k-1}^4\right)=\frac{1}{4}(x_n^4-x_0^4)=\frac{1}{4}.$$
But how to make the inequality more precisely?
Best Answer
For every partition $0 = x_0 \le x_1 \le x_2 \le \cdots \le x_n = 1$ of $[0, 1]$ is $$ \sum_{k=1}^n x_k^3(x_k - x_{k-1}) \ge \frac 14 + \frac 38 \frac 1n $$ and the factor $\boxed{\lambda = \frac 38}$ is best possible.
Proof: The identity $$ a^3 (a-b) = \frac 14 (a^4 - b^4) + \frac 38 (a^2-b^2)^2 + \frac 18 (a-b)^3(3a+b) $$ can be verified easily. It follows that $$ S_n =\sum_{k=1}^n x_k^3(x_k - x_{k-1}) = A_n + B_n + C_n $$ where $$ A_n = \frac 14 \sum_{k=1}^n (x_k^4 - x_{k-1}^4) = \frac 14 \, , $$ then, using the Cauchy-Schwarz inequality, $$ B_n = \frac 38 \sum_{k=1}^n (x_k^2 - x_{k-1}^2)^2 \underset{(*)}{\ge} \frac 38 \frac 1n \left( \sum_{k=1}^n (x_k^2 - x_{k-1}^2)\right)^2 = \frac{3}{8n} \, , $$ and finally, $$ C_n = \frac 18 \sum_{k=1}^n (x_k-x_{k-1})^3(3 x_k + x_{k-1}) \ge 0 \, . $$ Adding these inequalities gives the desired estimate.
In order to show that the factor $3/8$ is best possible we consider the partitions $x_k = \sqrt{k/n}$, $0 \le k \le n$. Then equality holds at $(*)$ in the estimate of $B_n$, so that $$ S_n = \frac 14 + \frac 38 \frac 1n + C_n \, . $$ Now $$ 0 \le C_n \le \frac 12 \sum_{k=1}^n \left( \frac kn \right)^{1/2} \left( \left(\frac kn \right)^{1/2} - \left(\frac {k-1}n \right)^{1/2}\right)^{3} \\ = \frac{1}{2n^2} \sum_{k=1}^n k^2 \left( 1 - \left( 1 - \frac 1k \right)^{1/2}\right)^{3} \, . $$ Using $\sqrt{1-x} \ge 1-x$ for $0 \le x \le 1$ it follows that $$ 0 \le C_n \le \frac{1}{2n^2} \sum_{k=1}^n \frac 1k \le \frac{1 + \ln(n)}{2 n^2} \, . $$ We have therefore shown that for these partitions $$ S_n = \frac 14 + \frac{3}{8n} + o\left( \frac{1}{n}\right) $$ and that concludes the proof.