Croatian Mathematical Competitions

The quadruple $(a, b, c, d) = (1, 1, 1, 1)$ satisfies the given condition for every positive integer $k$, which means that $D_k \ge 4$ for every positive integer $k$. We will prove that $D_k = 4$ for every $k \ge 2$. If $k \ge 2$, the power mean inequality gives us:
$$\sqrt{\frac{a^2 + b^2 + c^2 + d^2}{4}} \le \sqrt[k]{\frac{a^k + b^k + c^k + d^k}{4}} = 1,$$
i.e.
$$a^2 + b^2 + c^2 + d^2 \le 4.$$
By introducing the substitution $x = a^2$, $y = b^2$, $z = c^2$, $w = d^2$ the previous inequality turns into
$$x + y + z + w \le 4,$$
and the given inequality turns into
$$xyz + xyw + xzw + yzw \le 4. \qquad (11)$$
Obviously, it is sufficient to prove the inequality:
$$16(xyz + xyw + xzw + yzw) \le (x + y + z + w)^3.$$
We have
$$(x - y + z - w)^2 \ge 0,$$
which can be equivalently written as
$$(x + y + z + w)^2 \ge 4(xy + yz + zw + wx). \qquad (12)$$
We also have
$$(x + y + z + w)(xy + yz + zw + wx) \ge 4(xyz + xyw + xzw + yzw). \qquad (13)$$
Namely, the last inequality is equivalent to
$$(x^2 + z^2)(y + w) + (y^2 + w^2)(x + z) \ge 2(xyz + xzw + xyw + yzw)$$
which holds by the AM-GM inequality
$$x^2 + z^2 \ge 2xz \quad \text{and} \quad y^2 + w^2 \ge 2yw.$$
Multiplying (12) and (13) gives us (11).

If $k = 1$ then $D_k > 4$. Namely, plugging the quadruple $(\frac{4}{3}, \frac{4}{3}, \frac{4}{3}, 0)$ gives us $D_1 \ge (\frac{4}{3})^6 > 4$. We will prove that $D_1$ is precisely equal to $(\frac{4}{3})^6$.

Let us assume that $a, b, c, d$ are real numbers such that $a + b + c + d = 1$ and $a \ge b \ge c \ge d \ge 0$. We denote $a' = a, b' = b + \frac{d}{2}, c' = c + \frac{d}{2}, d' = 0$. Replacing $(a, b, c, d) \rightarrow (a', b', c', d')$ the left hand side of the given inequality increases. Namely,
$$\begin{aligned} & (a'b'c')^2 + (a'b'd')^2 + (a'c'd')^2 + (b'c'd')^2 \\ &= a^2 \left(b + \frac{d}{2}\right)^2 \left(c + \frac{d}{2}\right)^2 \\ &= a^2 \left(b^2 + bd + \frac{d^2}{4}\right) \left(c^2 + cd + \frac{d^2}{4}\right) \\ &\ge a^2 \cdot b^2 \cdot c^2 + a^2 \cdot b^2 \cdot cd + a^2 \cdot bd \cdot c^2 + a^2 \cdot bd \cdot cd \\ &\ge (abc)^2 + (abd)^2 + (acd)^2 + (bcd)^2, \end{aligned}$$
where the last inequality holds because of the ordering of numbers $a, b, c, d$. This means that the left hand side of the inequality attains its maximum when one of the numbers is equal to zero, so without loss of generality we can assume that $d = 0$. Now it follows that:
$$(abc)^2 + (abd)^2 + (acd)^2 + (bcd)^2 = (abc)^2 \stackrel{\text{A-G}}{\le} \left(\frac{a+b+c}{3}\right)^6 = \left(\frac{4}{3}\right)^6,$$
which means that $D_1 = \left(\frac{4}{3}\right)^6$.