三角不等式

Let $n$ be a natural number and let $0\lt x\lt{\pi}$. Then, here are my questions. Question 1: Is the following true? $$\sum_{k=1}^{n}\frac{\cos(kx)}{k}\gt -1$$ Question 2: Is the following true? $$\sum_{k=1}^{n}\frac{\sin(kx)}{k}\gt0$$ This is a possible hint for solution; perhaps someone can finish it along these lines (it won't fit as a comment). We have $$\sin x+\dfrac{\sin 2x}{2}+\dfrac{\sin 3x}{3}+\ldots+ \dfrac{\sin nx}{n}=\sum_{k=1}^n\int_0^x\cos kt\,dt,$$ $$2\sum_{k=1}^n\cos kt=\sin((n+1/2)t)/\sin(t/2)-1$$ (by taking the real part of $\sum_{k=1}^n e^{ikt}$) so we want to show $$\int_0

形式 \[ \sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2 \geq \sum_{i=1}^{n}a_i^{2}b_i^2 \] 等号成立的条件： \[ iff:b_i=0 || \exists k \in \mathbb {R},a_i=k \cdot b_i(i \in \mathbb{N^+}) \] 证明 法一：参数配方 思路：巧妙的把常数与方程结合起来，利用性质即可。 证明： 构造函数: \[ f(t)=\sum_{i=1}^{n}b_i^2\cdot t^2-2\sum_{i=1}^{n}a_ib_it+\sum_{i=1}^{n}a_i^2 \] 化简函数： \[ f(t)=\sum_{i=1}^{n}b_i^2\cdot t^2-2\sum_{i=1}^{n}a_ib_it+\sum_{i=1}^{n}a_i^2 \] \[ =\sum_{i=1}^{n}(b_i^2t^2-2a_ib_it+a_i^2) \] \[ =\sum_{i=1}^{n}(b_i^2t^2+a_i^2-2a_ib_it) \] \[ =\sum_{i=1}^{n}(b_it-a_i)^2 \] 所以： \[ f(t) \geq 0 \] \[ \Delta t=b^2-4ac \] \[ =4\sum_{i=1}^{n}a_i^2b_i^2-4\times