contest-mathinequalitytrigonometry
I am solving a question and I can't get over this step: proving $$\sin \frac{1}{k} > \frac{1}{k} – \frac{1}{k^2}$$
where $k$ is a positive integer.
I tried using induction, but I failed. One of my friends managed to prove it using derivatives but I am searching for a solution which does not involve calculus or series. Proving this would help me solve a convergence problem.
The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$ \cos(x)\le\frac{\sin(x)}{x}\le1\tag{2} $$ Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3} $$ Also, dividing $(2)$ by $\cos(x)$, we get that $$ 1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4} $$ Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$
Knowing the answer it's not that difficult to get it another way. Let $y = \cos 2x$. We have $\sin^2 x = \frac{1-y}{2}$, $\cos^2 x =\frac{1+y}{2}$. Then \begin{align} \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 &= \left(\frac{1-y}{2} + \frac{2}{1-y}\right)^2 + \left(\frac{1+y}{2} + \frac{2}{1+y}\right)^2 = \\ &= \frac{y^6+7y^4-y^2+25}{2(1-y^2)^2} = \\ &= \frac{25}{2} + \frac{y^2(y^4-18y^2+49)}{2(1-y^2)^2} = \\ &= \frac{25}{2} + y^2\Big(\frac{1}{2} + \frac{8}{1-y^2} + \frac{16}{(1-y^2)^2}\Big) \end{align} Since $y^2 \le 1$, the expression in the brackets is strictly positive.
Best Answer
We just have to think about:$$f(x)=\sin x-x+x^2,x\in[0,1].\\$$ $$f'(x)=\cos x-1+2x.\\$$Let$\ g(x)=f'(x),$than$$g'(x)=-\sin x+2.\\$$ When $x\in[0,1]$ $$g'(x)>0.\\$$At this point, $f'(x)(=g(x))$ is monotonically increasing.
So$$f'(x)\geq f'(0)=0,$$if and only if $x=0$,the equal sign holds.
Than we can know $f(x)$ is monotonically increasing.
So$$f(x)\geq f(0)=0,$$if and only if $x=0$,the equal sign holds.
Let x=$\frac{1}{k}>0$,come to the conclusion:$$\sin \frac{1}{k}>\frac{1}{k}-\frac{1}{k^2}.$$