# Ngô Quốc Anh

## October 15, 2007

### Jackson’s inequality

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 14:28

Prove that with $n$ be a possitive integer and $x\in(0,\pi)$ $\displaystyle \sin x + \frac{1}{2}\sin 2x + ... + \frac{1}{n}\sin nx > 0$.

Solution. For $n=1$ statement is correct. Consider $n>1$. If some of the sums $\displaystyle \frac{1}{k}\sin kx+\frac{1}{k+1}\sin (k+1)x+...+\frac{1}{n}\sin nx\geq 0$

( $k=2,...,n$) then we are done, since in this case, due to induction proposition, whole sum is $\displaystyle S(x)=\left(\sin x+...+\frac{1}{k-1}\sin (k-1)x\right)+\left(\frac{1}{k}\sin kx+...+\frac{1}{n}\sin nx\right)=(>0)+(\geq 0)$.

Now suppose all mentioned sums are negative and whole sum is non-positive. Then sum them up $\displaystyle \sin x+\sin 2x+...+\sin nx<0$.

Since $S(t)$ is positive in neighbourhood of $\pi$ and $S(x)\leq 0$, we may WLOG assume $S'(x)\geq 0$ (just consider the largest $x$ with condition $S(x)\leq 0$). Thus $\displaystyle\cos x+\cos 2x+...+\cos nx\geq 0$.

Using well known formulas for $\sum \sin kx$ and $\sum \cos kx$ we obtain $\displaystyle\cos(x/2)-\cos(n+1/2)x<0$

and $\displaystyle\sin(n+1/2)x-\sin(x/2)\geq 0$.

It immediatelly implies $\displaystyle 1=\cos^2(n+1/2)x+\sin^2(n+1/2)x>\cos^2(x/2)+\sin^2(x/2)=1$