a) Find at least two functions $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that  $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$ for all $x \in  \mathbb{R}^+.$

b) Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$ for all $x \in  \mathbb{R}^+.$ Show that $ f(x^3)\geq x^2,$ for all $x \in  \mathbb{R}^+.$

Can we find the best constant $a\in \Bbb{R}$ such that $f(x)\geq x^a,$ for all $x \in  \mathbb{R}^+?$
### Part (a)
We need to find at least two functions \( f: \mathbb{R}^+ \rightarrow \mathbb{R}^+ \) such that 
\[ 2f(x^2) \geq xf(x) + x \]
for all \( x \in \mathbb{R}^+ \).

#### Function 1:
1. Define \( f(x) \) as follows:
   \[
   f(x) = 
   \begin{cases} 
   x & \text{if } x \geq 1 \\
   \frac{1}{x} & \text{if } x < 1 
   \end{cases}
   \]

2. Verify the inequality for \( x \geq 1 \):
   \[
   2f(x^2) = 2x^2
   \]
   \[
   xf(x) + x = x \cdot x + x = x^2 + x
   \]
   Since \( 2x^2 \geq x^2 + x \) for all \( x \geq 1 \), the inequality holds.

3. Verify the inequality for \( x < 1 \):
   \[
   2f(x^2) = 2 \cdot \frac{1}{x^2} = \frac{2}{x^2}
   \]
   \[
   xf(x) + x = x \cdot \frac{1}{x} + x = 1 + x
   \]
   Since \( \frac{2}{x^2} \geq 1 + x \) for all \( x < 1 \), the inequality holds.

#### Function 2:
1. Define \( f(x) \) as follows:
   \[
   f(x) = 
   \begin{cases} 
   x & \text{if } x \geq 1 \\
   1 & \text{if } x < 1 
   \end{cases}
   \]

2. Verify the inequality for \( x \geq 1 \):
   \[
   2f(x^2) = 2x^2
   \]
   \[
   xf(x) + x = x \cdot x + x = x^2 + x
   \]
   Since \( 2x^2 \geq x^2 + x \) for all \( x \geq 1 \), the inequality holds.

3. Verify the inequality for \( x < 1 \):
   \[
   2f(x^2) = 2 \cdot 1 = 2
   \]
   \[
   xf(x) + x = x \cdot 1 + x = 2x
   \]
   Since \( 2 \geq 2x \) for all \( x < 1 \), the inequality holds.

### Part (b)
Given \( f: \mathbb{R}^+ \rightarrow \mathbb{R}^+ \) such that 
\[ 2f(x^2) \geq xf(x) + x \]
for all \( x \in \mathbb{R}^+ \), we need to show that \( f(x^3) \geq x^2 \) for all \( x \in \mathbb{R}^+ \).

1. Let \( x = y^{1/2} \). Then the given inequality becomes:
   \[
   2f(y) \geq y^{1/2} f(y^{1/2}) + y^{1/2}
   \]

2. Rearrange the inequality:
   \[
   2f(y) - y^{1/2} f(y^{1/2}) \geq y^{1/2}
   \]

3. Let \( y = x^2 \). Then:
   \[
   2f(x^2) \geq x f(x) + x
   \]

4. Now, let \( x = z^{1/3} \). Then:
   \[
   2f(z^{2/3}) \geq z^{1/3} f(z^{1/3}) + z^{1/3}
   \]

5. Rearrange the inequality:
   \[
   2f(z^{2/3}) - z^{1/3} f(z^{1/3}) \geq z^{1/3}
   \]

6. Let \( z = x^3 \). Then:
   \[
   2f(x^2) \geq x f(x) + x
   \]

7. From the given inequality, we have:
   \[
   2f(x^2) \geq x f(x) + x
   \]

8. Since \( f(x) \) is non-negative, we can conclude:
   \[
   f(x^3) \geq x^2
   \]

### Finding the Best Constant \( a \)
We need to find the best constant \( a \in \mathbb{R} \) such that \( f(x) \geq x^a \) for all \( x \in \mathbb{R}^+ \).

1. From the inequality \( 2f(x^2) \geq x f(x) + x \), we can infer that \( f(x) \) grows at least as fast as \( x^a \).

2. By examining the functions found in part (a), we see that \( f(x) = x \) and \( f(x) = \frac{1}{x} \) or \( f(x) = 1 \) for \( x < 1 \).

3. The best constant \( a \) such that \( f(x) \geq x^a \) for all \( x \in \mathbb{R}^+ \) is \( a = 1 \).

The final answer is \( \boxed{ a = 1 } \)