Prove that for all real numbers $ a,b$ with $ ab>0$ we have:

$ \sqrt[3]{\frac{a^2 b^2 (a\plus{}b)^2}{4}} \le \frac{a^2\plus{}10ab\plus{}b^2}{12}$

and find the cases of equality. Hence, or otherwise, prove that for all real numbers $ a,b$

$ \sqrt[3]{\frac{a^2 b^2 (a\plus{}b)^2}{4}} \le \frac{a^2\plus{}ab\plus{}b^2}{3}$

and find the cases of equality.
1. **First Inequality:**
   We need to prove that for all real numbers \(a, b\) with \(ab > 0\), the following inequality holds:
   \[
   \sqrt[3]{\frac{a^2 b^2 (a+b)^2}{4}} \le \frac{a^2 + 10ab + b^2}{12}
   \]

   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that for any non-negative real numbers \(x_1, x_2, \ldots, x_n\),
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   with equality if and only if \(x_1 = x_2 = \cdots = x_n\).

   Let's apply the AM-GM inequality to the terms \(ab, ab, \frac{(a+b)^2}{4}\):
   \[
   \frac{ab + ab + \frac{(a+b)^2}{4}}{3} \ge \sqrt[3]{ab \cdot ab \cdot \frac{(a+b)^2}{4}}
   \]
   Simplifying the right-hand side:
   \[
   \sqrt[3]{ab \cdot ab \cdot \frac{(a+b)^2}{4}} = \sqrt[3]{\frac{a^2 b^2 (a+b)^2}{4}}
   \]

   Therefore, we have:
   \[
   \frac{ab + ab + \frac{(a+b)^2}{4}}{3} \ge \sqrt[3]{\frac{a^2 b^2 (a+b)^2}{4}}
   \]

   Now, let's simplify the left-hand side:
   \[
   \frac{ab + ab + \frac{(a+b)^2}{4}}{3} = \frac{2ab + \frac{a^2 + 2ab + b^2}{4}}{3} = \frac{2ab + \frac{a^2 + 2ab + b^2}{4}}{3}
   \]
   \[
   = \frac{8ab + a^2 + 2ab + b^2}{12} = \frac{a^2 + 10ab + b^2}{12}
   \]

   Thus, we have shown that:
   \[
   \sqrt[3]{\frac{a^2 b^2 (a+b)^2}{4}} \le \frac{a^2 + 10ab + b^2}{12}
   \]

   Equality holds if and only if \(ab = ab = \frac{(a+b)^2}{4}\), which implies \(a = b\).

2. **Second Inequality:**
   We need to prove that for all real numbers \(a, b\),
   \[
   \sqrt[3]{\frac{a^2 b^2 (a+b)^2}{4}} \le \frac{a^2 + ab + b^2}{3}
   \]

   From the first part, we already have:
   \[
   \sqrt[3]{\frac{a^2 b^2 (a+b)^2}{4}} \le \frac{a^2 + 10ab + b^2}{12}
   \]

   Now, we need to show that:
   \[
   \frac{a^2 + 10ab + b^2}{12} \le \frac{a^2 + ab + b^2}{3}
   \]

   Simplifying the right-hand side:
   \[
   \frac{a^2 + ab + b^2}{3} = \frac{4(a^2 + ab + b^2)}{12}
   \]

   Therefore, we need to show:
   \[
   a^2 + 10ab + b^2 \le 4(a^2 + ab + b^2)
   \]

   Simplifying the inequality:
   \[
   a^2 + 10ab + b^2 \le 4a^2 + 4ab + 4b^2
   \]
   \[
   0 \le 3a^2 - 6ab + 3b^2
   \]
   \[
   0 \le 3(a^2 - 2ab + b^2)
   \]
   \[
   0 \le 3(a - b)^2
   \]

   Since \((a - b)^2 \ge 0\) for all real numbers \(a\) and \(b\), the inequality holds.

   Equality holds if and only if \(a = b\).

The final answer is \( \boxed{ \sqrt[3]{\frac{a^2 b^2 (a+b)^2}{4}} \le \frac{a^2 + ab + b^2}{3} } \) with equality when \(a = b\).