Can you please clarify how to prove this?
$$\int_{0}^{\pi}\left(\frac{1+\cos t}{2}\right)^{k}\operatorname{d}t>\int_{0}^{\pi}\left(\frac{1+\cos t}{2}\right)^{k}\sin t\operatorname{d}t$$
Here, $k=1,2,3,...$
This is from Section 4.24 in Rudin Real and Complex Analysis. This inequality is part of the proof to show that there is a trigonometric Polynomial approximation for all continuous complex periodic functions.
My question is about showing that the condition is greater than ">" as opposed to ">=".
I am able to plot these functions and sort of intuitively see this as shown below.
note that if $f$ is continous on $[a,b]$ then the integral of f on the open interval $]a,b[$ have the same value of the integral on the closed interval $[a,b]$.You can see the the proof on this link Integrating on open vs. closed intervals .Regarding your question , $\forall x \in(0,\pi)$ , $0<sinx<1$, and $1+\cos(x)>0$ , so $\forall x \in(0,\pi),$ $(\frac{1+\cos(x)}{2})^k>(\frac{1+\cos(x)}{2})^k$$\sin(x)$ then $\int_{0}^{\pi}(\frac{1+\cos(x)}{2})^kdx>\int_{0}^{\pi}(\frac{1+\cos(x)}{2})^k\ sinxdx$