Can You Prove this Trigonometric Inequality?

[anemone]

If $x\in \left(0,\,\dfrac{\pi}{2}\right)$, $0\le a \le b$ and $0\le c \le 1$, prove that $\left(\dfrac{c+\cos x}{c+1}\right)^b<\left(\dfrac{\sin x}{x}\right)^a$.

 

[jvicens]

Thank you for your post. I would like to provide a proof for the statement you have proposed.

First, let us rewrite the expression on the left side of the inequality:

$\left(\dfrac{c+\cos x}{c+1}\right)^b = \left(1+\dfrac{\cos x-1}{c+1}\right)^b$

Since $x\in \left(0,\,\dfrac{\pi}{2}\right)$, we know that $\cos x < 1$. Therefore, $\dfrac{\cos x-1}{c+1}$ is negative for all values of $c$ and $x$ in the given range.

Next, we can use the well-known inequality $1+x < e^x$ for all $x<0$. Applying this to our expression, we have:

$\left(1+\dfrac{\cos x-1}{c+1}\right)^b < e^{\frac{\cos x-1}{c+1}\cdot b}$

Now, since $0\le a\le b$, we can also say that $\dfrac{a}{b}\le 1$. Using this, we can rewrite the expression on the right side of the inequality as:

$\left(\dfrac{\sin x}{x}\right)^a = \left(1+\dfrac{\sin x-x}{x}\right)^a$

Again, since $x\in \left(0,\,\dfrac{\pi}{2}\right)$, we know that $\sin x < x$. Therefore, $\dfrac{\sin x-x}{x}$ is negative for all values of $x$ in the given range.

Using the same inequality as before, we have:

$\left(1+\dfrac{\sin x-x}{x}\right)^a < e^{\frac{\sin x-x}{x}\cdot a}$

Now, we can see that the only difference between the two expressions is the exponent. Since $a\le b$, we know that $\dfrac{a}{c+1}\le \dfrac{b}{c+1}$. Therefore, the exponent in the first expression is smaller than the exponent in the second expression.

Combining all of these inequalities, we can conclude that:

$\left(\dfrac{c+\cos x}{c+1}\right)^b < e^