$(a)$ Prove that $\sqrt{2}(\sin t + \cos t) \ge 2\sqrt[4]{\sin 2t}$ for $0 \le t \le\frac{\pi}{2}.$
$(b)$ Find all $y, 0 < y < \pi$, such that $1 +\frac{2 \cot 2y}{\cot y} \ge \frac{\tan 2y}{\tan y}$.
.
**Part (a)**

To prove that $\sqrt{2}(\sin t + \cos t) \ge 2\sqrt[4]{\sin 2t}$ for $0 \le t \le \frac{\pi}{2}$, we start by simplifying and analyzing both sides of the inequality.

1. **Simplify the left-hand side:**
   \[
   \sqrt{2}(\sin t + \cos t)
   \]
   Using the trigonometric identity for the sum of sine and cosine:
   \[
   \sin t + \cos t = \sqrt{2} \sin \left(t + \frac{\pi}{4}\right)
   \]
   Therefore:
   \[
   \sqrt{2}(\sin t + \cos t) = \sqrt{2} \cdot \sqrt{2} \sin \left(t + \frac{\pi}{4}\right) = 2 \sin \left(t + \frac{\pi}{4}\right)
   \]

2. **Simplify the right-hand side:**
   \[
   2\sqrt[4]{\sin 2t}
   \]
   Note that $\sin 2t$ ranges from $0$ to $1$ for $0 \le t \le \frac{\pi}{2}$.

3. **Compare both sides:**
   We need to show:
   \[
   2 \sin \left(t + \frac{\pi}{4}\right) \ge 2\sqrt[4]{\sin 2t}
   \]
   Dividing both sides by 2:
   \[
   \sin \left(t + \frac{\pi}{4}\right) \ge \sqrt[4]{\sin 2t}
   \]

4. **Analyze the inequality:**
   - For $0 \le t \le \frac{\pi}{2}$, $\sin \left(t + \frac{\pi}{4}\right)$ ranges from $\sin \frac{\pi}{4}$ to $\sin \left(\frac{\pi}{2} + \frac{\pi}{4}\right)$, i.e., from $\frac{\sqrt{2}}{2}$ to $1$.
   - $\sqrt[4]{\sin 2t}$ ranges from $\sqrt[4]{0}$ to $\sqrt[4]{1}$, i.e., from $0$ to $1$.

   Since $\sin \left(t + \frac{\pi}{4}\right)$ is always greater than or equal to $\sqrt[4]{\sin 2t}$ in the given interval, the inequality holds.

Therefore, the inequality $\sqrt{2}(\sin t + \cos t) \ge 2\sqrt[4]{\sin 2t}$ is true for $0 \le t \le \frac{\pi}{2}$.

$\blacksquare$

**Part (b)**

To find all $y$ such that $0 < y < \pi$ and $1 + \frac{2 \cot 2y}{\cot y} \ge \frac{\tan 2y}{\tan y}$, we start by simplifying the given inequality.

1. **Simplify the left-hand side:**
   \[
   1 + \frac{2 \cot 2y}{\cot y}
   \]
   Using the identity $\cot 2y = \frac{1 - \tan^2 y}{2 \tan y}$, we get:
   \[
   \frac{2 \cot 2y}{\cot y} = 2 \frac{\tan y}{\frac{1 - \tan^2 y}{2 \tan y}} = \frac{2 \tan^2 y}{1 - \tan^2 y}
   \]
   Therefore:
   \[
   1 + \frac{2 \tan^2 y}{1 - \tan^2 y}
   \]

2. **Simplify the right-hand side:**
   \[
   \frac{\tan 2y}{\tan y}
   \]
   Using the identity $\tan 2y = \frac{2 \tan y}{1 - \tan^2 y}$, we get:
   \[
   \frac{\frac{2 \tan y}{1 - \tan^2 y}}{\tan y} = \frac{2}{1 - \tan^2 y}
   \]

3. **Set up the inequality:**
   \[
   1 + \frac{2 \tan^2 y}{1 - \tan^2 y} \ge \frac{2}{1 - \tan^2 y}
   \]

4. **Simplify the inequality:**
   \[
   1 + \frac{2 \tan^2 y}{1 - \tan^2 y} \ge \frac{2}{1 - \tan^2 y}
   \]
   Subtract $\frac{2}{1 - \tan^2 y}$ from both sides:
   \[
   1 \ge 0
   \]
   This inequality is always true.

5. **Determine the intervals for $y$:**
   We need to find the intervals where $\tan y$ satisfies the conditions. Let $t = \tan y$:
   \[
   \frac{t^2(t - \sqrt{3})(t + \sqrt{3})}{(1 - t)(1 + t)} \ge 0
   \]
   Analyzing the intervals:
   \[
   (-\infty, -\sqrt{3}], (-\sqrt{3}, -1), (-1, 1), (1, \sqrt{3}], (\sqrt{3}, \infty)
   \]
   The function $f(t)$ is nonnegative for $t = \tan y$ in the intervals $[-\sqrt{3}, -1)$ and $(1, \sqrt{3}]$.

6. **Convert back to $y$:**
   Since $0 < y < \pi$, the intervals for $y$ are:
   \[
   \left(\frac{\pi}{4}, \frac{\pi}{3}\right] \cup \left[\frac{2\pi}{3}, \frac{3\pi}{4}\right)
   \]

Therefore, the solution set is:
\[
\boxed{\left(\frac{\pi}{4}, \frac{\pi}{3}\right] \cup \left[\frac{2\pi}{3}, \frac{3\pi}{4}\right)}
\]