# A delicate trigonometric inequality

I would like to show that for $u\in[0,\pi/2]$ we have $$\frac{u}{\sin(u)}+\frac{1}{4}\frac{u-\sin(u)\cos(u)}{u-\sin(u)} \geq 2$$ Any idea?

• for $u=0$ we have a problem in the inequality – Dr. Sonnhard Graubner Nov 30 '16 at 16:06
• No near zero if you use power series you have the inequality. But I did not find how to prove it on the interval $[0,\pi/2]$. – Babyblog Nov 30 '16 at 16:09
• You can try taking the derivative and showing that it's strictly increasing on the open interval. However that will be quite messy. – lordoftheshadows Nov 30 '16 at 16:10
• @Babyblog it's defined everywhere on the interval except 0. That is a removable discontinuity though. It also doesn't really matter though because the limit is $2$ there if I did my math correctly. – lordoftheshadows Nov 30 '16 at 16:11
• I tried this but it is not easy to prove that the derivative is positive... I was not able to prove it in fact. – Babyblog Nov 30 '16 at 16:12

Two inequalities. The first one

$$\frac{u}{\sin u}\geq 1+\frac{u^2}{6}$$

is easy to prove using power series. The second one

$$1-\cos u\geq \frac{u^2}{2}-\frac{u^4}{24}\geq \frac{u-\sin u}{\sin u}\left(3-\frac{3u^2}{5}\right)$$

can also be derived using power series and can be rewritten as

$$\frac{\sin u(1-\cos u)}{u-\sin u}\geq 3-\frac{3u^2}{5}$$ Multiply this last inequality by $1/4$ and add to the first inequality above to get

$$\frac{u}{\sin u}+\frac{1}{4}\frac{\sin u(1-\cos u)}{u-\sin u}\geq \frac{7}{4}+\frac{u^2}{60}\geq \frac{7}{4}$$

Now add $\frac{1}{4}$ to both sides and simplify

$$\frac{u}{\sin u}+\frac{1}{4}\frac{u-\sin u \cos u}{u-\sin u}\geq 2$$

• I am ok for $u$ near zero. I can check this on matlab, but I can't prove your inequalities with power series on all the interval $[0,\pi/2]$. – Babyblog Dec 5 '16 at 18:08
• I don't understand why it's true on $[0,\pi/2]$. – Babyblog Dec 5 '16 at 18:09
• @Babyblog I was referring to inequalities such as $\sin u \leq u$ and $\sin u \leq u-\frac{u^3}{6}+\frac{u^5}{120}$ which are valid in the interval $[0,\pi /2]$. If we multiply this second inequality by $1+\frac{u^2}{6}$ we can see that $\left(1+\frac{u^2}{6}\right)\sin u \leq u$ – Lozenges Dec 6 '16 at 15:04