Problem 1787

### First Inequality:
Given: $a, b, c, d$ are positive numbers such that $a + b + c + d = 3$. We need to prove:
$$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} \le \frac{1}{a^2b^2c^2d^2}$$

1. Define the function:
$$f(a, b, c, d) = a^2b^2c^2 + b^2c^2d^2 + c^2d^2a^2 + d^2a^2b^2$$
We need to show that $f(a, b, c, d) \le 1$ for $a + b + c + d = 3$.

2. Compactness and Maximum:
The set $\{(a, b, c, d) \in \mathbb{R}^4 \mid a + b + c + d = 3\} \cap [0, 1]^4$ is compact, so $f(a, b, c, d)$ attains a maximum over it.

3. Symmetry and Critical Points:
Suppose we fix $c$ and $d$ and let $a, b$ vary with $a + b = t$ fixed. We claim that $f(a, b, c, d)$ is maximized either when $a = b = t/2$ or one of $a, b$ is 0 (so the other equals $t$).

4. Rewrite the function:
$$f(a, b, c, d) = (ab)^2(c^2 + d^2) + (a^2 + b^2)c^2d^2 = (ab)^2(c^2 + d^2) + (t^2 - 2ab)c^2d^2$$
Since $t, c, d$ are fixed, we want to maximize $(c^2 + d^2)(ab)^2 - 2cd(ab)$.

5. **Quadratic in $ab$**:
This is a quadratic in $ab$ with a positive leading coefficient, so it is maximized at an endpoint. Thus, we either minimize $ab$ (when one of $a, b$ is 0) or maximize it (when $a = b$).

6. Cases:
- If two or more variables are zero, $f(a, b, c, d) = 0$.
- If one variable is zero, say $d = 0$, then $f(a, b, c, d) = a^2b^2c^2 = 1$.
- If no variables are zero, then $a = b = c = d = \frac{3}{4}$, and:
$$f(a, b, c, d) = \left(\frac{3}{4}\right)^6 = \frac{729}{4096} < 1$$

Therefore, $f(a, b, c, d) \le 1$ for all nonnegative reals $a, b, c, d$ with $a + b + c + d = 3$.

$\blacksquare$

### Second Inequality:
Given: $a, b, c, d$ are positive numbers such that $a + b + c + d = 3$. We need to prove:
$$\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} + \frac{1}{d^3} \le \frac{1}{a^3b^3c^3d^3}$$

1. Define the function:
$$g(a, b, c, d) = a^3b^3c^3 + b^3c^3d^3 + c^3d^3a^3 + d^3a^3b^3$$
We need to show that $g(a, b, c, d) \le 1$ for $a + b + c + d = 3$.

2. Compactness and Maximum:
The set $\{(a, b, c, d) \in \mathbb{R}^4 \mid a + b + c + d = 3\} \cap [0, 1]^4$ is compact, so $g(a, b, c, d)$ attains a maximum over it.

3. Symmetry and Critical Points:
Suppose we fix $c$ and $d$ and let $a, b$ vary with $a + b = t$ fixed. We claim that $g(a, b, c, d)$ is maximized either when $a = b = t/2$ or one of $a, b$ is 0 (so the other equals $t$).

4. Rewrite the function:
$$g(a, b, c, d) = (ab)^3(c^3 + d^3) + (a^3 + b^3)c^3d^3 = (ab)^3(c^3 + d^3) - (t^3 - 3t(ab))(c^3d^3)$$
Since $t, c, d$ are fixed, we want to maximize $(c^3 + d^3)(ab)^3 - 3tc^3d^3(ab)$.

5. **Cubic in $ab$**:
This is a cubic in $ab$ with a positive leading coefficient, so it is maximized at an endpoint. Thus, we either minimize $ab$ (when one of $a, b$ is 0) or maximize it (when $a = b$).

6. Cases:
- If two or more variables are zero, $g(a, b, c, d) = 0$.
- If one variable is zero, say $d = 0$, then $g(a, b, c, d) = a^3b^3c^3 = 1$.
- If no variables are zero, then $a = b = c = d = \frac{3}{4}$, and:
$$g(a, b, c, d) = \left(\frac{3}{4}\right)^9 = \frac{19683}{262144} < 1$$

Therefore, $g(a, b, c, d) \le 1$ for all nonnegative reals $a, b, c, d$ with $a + b + c + d = 3$.

$\blacksquare$