# Another inequality with exponential sum and trigonometry

Playing with $$\tan$$ and $$\sin$$ I get :

Let $$a,b>0$$ such that $$a+b=1$$ then we have : $$\Big(\sin\Big(\frac{a\pi}{2}\Big)\Big)^{2\tan\Big(\frac{b\pi}{2}\Big)}+\Big(\sin\Big(\frac{b\pi}{2}\Big)\Big)^{2\tan\Big(\frac{a\pi}{2}\Big)}\leq 1$$

I try to exploit the derivative and use the closed interval method . Much harder I try to, use power series at $$x=0.5$$

Another way will be to divide the problem in two as:

$$\Big(\sin\Big(\frac{x\pi}{2}\Big)\Big)^{2\tan\Big(\frac{(1-x)\pi}{2}\Big)}\leq 0.5+\alpha(0.5x-x^2)+\beta(0.5-x)$$

And

$$\Big(\sin\Big(\frac{(1-x)\pi}{2}\Big)\Big)^{2\tan\Big(\frac{(x)\pi}{2}\Big)}\leq 0.5-\alpha(0.5^2-x0.5)-\beta(0.5-x)$$ I can find the right constants but I can't prove it .

Finally I think we can do something with the Bernoulli's inequality .

The equality case is : $$a=b=0.5$$ and $$a=0$$ or $$a=1$$

Thanks a lot for your comment and answer .

Due to symmetry, assume that $$a \ge b$$. By using $$a = 1- b$$ and $$b\in (0, \frac{1}{2}]$$, the desired inequality is written as $$(\sin \tfrac{(1-b)\pi}{2})^{2\tan \frac{b\pi}{2}} + (\sin \tfrac{b\pi}{2})^{2\tan \frac{(1-b)\pi}{2}} \le 1.$$

Let us first prove the case when $$b\in (0, \frac{1}{3}]$$.

We give the following auxiliary results (Facts 1 through 5). Their proof is not difficult and hence omitted.

Fact 1: $$1 - \frac{7}{6} b^2 \ge \sin \frac{(1-b)\pi}{2}$$ for $$b\in (0, \frac{1}{3}]$$.

Fact 2: $$\tan \frac{b\pi}{2} \ge \frac{b\pi}{2} \ge \frac{3}{2}b$$ for $$b \in (0, \frac{1}{2}]$$.

Fact 3: $$\sin \frac{b\pi}{2} \le \frac{b\pi}{2} \le \frac{11}{7}b$$ for $$b \in (0, \frac{1}{2}]$$.

Fact 4: $$\tan \frac{(1-b)\pi}{2} \ge \frac{19}{5}-\frac{25}{4}b$$ for $$b \in (0, \frac{1}{3}]$$.

Fact 5: $$(1-x)^r \le 1 - rx$$ for $$r \in (0, 1]$$ and $$0 < x < 1$$.

From Facts 1, 2 and 5, we have $$(\sin \tfrac{(1-b)\pi}{2})^{2\tan \frac{b\pi}{2}} \le (1 - \tfrac{7}{6}b^2)^{2\tan \frac{b\pi}{2}} \le (1 - \tfrac{7}{6}b^2)^{3b} \le 1 - \tfrac{7}{2}b^3.$$ From Facts 3 and 4, we have $$(\sin \tfrac{b\pi}{2})^{2\tan \frac{(1-b)\pi}{2}} \le (\tfrac{11}{7}b)^{2\tan \frac{(1-b)\pi}{2}} \le (\tfrac{11}{7}b)^{38/5 - 25b/2}.$$

It suffices to prove that for $$b\in (0, \frac{1}{3}]$$, $$1 - \tfrac{7}{2}b^3 + (\tfrac{11}{7}b)^{38/5 - 25b/2} \le 1.$$ It is not difficult with computer. Omitted.

• I have a question: why it's more difficult on the part $b\in[\frac{1}{3},0.5]$?Or it's just a problem of tools ?Furthermore with my other way I think I can achieve the proof .But let me the time for that. Thanks (+1) Commented Feb 26, 2020 at 19:17
• @The.old.boy The equality occurs at $b=0$ and $b=1/2$. We need bounds for them respectively. Maybe it is not really more difficult. By using appropriate bounds, one may prove the other part. Commented Feb 27, 2020 at 1:35