## Task 1.

Depending on the natural number $k$, determine the smallest real number $D_{k}$ such that

$$
(a b c)^{2}+(b c d)^{2}+(c d a)^{2}+(d a b)^{2} \leqslant D_{k}
$$

for all non-negative real numbers $a, b, c, d$ for which $a^{k}+b^{k}+c^{k}+d^{k}=4$.
## Solution.

The quadruple $(a, b, c, d)=(1,1,1,1)$ satisfies the given condition for every natural number $k$, which means that $D_{k} \geqslant 4$ for every natural number $k$. Let's prove that $D_{k}=4$ for all $k \geqslant 2$. If $k \geqslant 2$, by the inequality between the power means of order $k$ and 2, we have:

$$
\sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \leqslant \sqrt[k]{\frac{a^{k}+b^{k}+c^{k}+d^{k}}{4}}=1
$$

or equivalently,

$$
a^{2}+b^{2}+c^{2}+d^{2} \leqslant 4
$$

By introducing the substitution $x=a^{2}, y=b^{2}, z=c^{2}, w=d^{2}$, the previous inequality becomes

$$
x+y+z+w \leqslant 4
$$

and the desired inequality is

$$
x y z+x y w+x z w+y z w \leqslant 4
$$

It suffices to prove the inequality:

$$
16(x y z+x y w+x z w+y z w) \leqslant(x+y+z+w)^{3}
$$

We have

$$
(x-y+z-w)^{2} \geqslant 0
$$

which can be equivalently written as

$$
(x+y+z+w)^{2} \geqslant 4(x y+y z+z w+w x)
$$

Also, we have

$$
(x+y+z+w)(x y+y z+z w+w x) \geqslant 4(x y z+x y w+x z w+y z w)
$$

Indeed, the last inequality is equivalent to the inequality

$$
\left(x^{2}+z^{2}\right)(y+w)+\left(y^{2}+w^{2}\right)(x+z) \geqslant 2(x y z+x z w+x y w+y z w)
$$

which holds because by the A-G inequality

$$
x^{2}+z^{2} \geqslant 2 x z \quad \text { and } \quad y^{2}+w^{2} \geqslant 2 y w
$$

Multiplying inequalities (14) and (15) gives us the desired inequality (13).

For the case $k=1$, we have $D_{k}>4$. Indeed, by substituting the quadruple $\left(\frac{3}{4}, \frac{3}{4}, \frac{3}{4}, 0\right)$, we get $D_{1} \geqslant\left(\frac{4}{3}\right)^{6}>4$. Let's prove that $D_{1}$ is exactly that number.

Assume that for some four real numbers $a, b, c, d$ we have $a+b+c+d=1$ and $a \geqslant b \geqslant c \geqslant d \geqslant 0$. Let $a^{\prime}=a, b^{\prime}=b+\frac{d}{2}, c^{\prime}=c+\frac{d}{2}, d^{\prime}=0$. By replacing $(a, b, c, d) \rightarrow\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)$, the expression on the left side of the given inequality increases. Indeed, we have

$$
\begin{aligned}
& \left(a^{\prime} b^{\prime} c^{\prime}\right)^{2}+\left(a^{\prime} b^{\prime} d^{\prime}\right)^{2}+\left(a^{\prime} c^{\prime} d^{\prime}\right)^{2}+\left(b^{\prime} c^{\prime} d^{\prime}\right)^{2} \\
& =a^{2}\left(b+\frac{d}{2}\right)^{2}\left(c+\frac{d}{2}\right)^{2} \\
& =a^{2}\left(b^{2}+b d+\frac{d^{2}}{4}\right)\left(c^{2}+c d+\frac{d^{2}}{4}\right) \\
& \geqslant a^{2} \cdot b^{2} \cdot c^{2}+a^{2} \cdot b^{2} \cdot c d+a^{2} \cdot b d \cdot c^{2}+a^{2} \cdot b d \cdot c d \\
& \geqslant(a b c)^{2}+(a b d)^{2}+(a c d)^{2}+(b c d)^{2}
\end{aligned}
$$

where the last inequality holds due to the ordering of the numbers $a, b, c, d$. From this, it follows that the left side of the given inequality achieves its maximum when one of the numbers is zero. Without loss of generality, we can assume that $d=0$. However, then we have:

$$
(a b c)^{2}+(a b d)^{2}+(a c d)^{2}+(b c d)^{2}=(a b c)^{2} \stackrel{\text { A-G }}{\leqslant}\left(\frac{a+b+c}{3}\right)^{6}=\left(\frac{4}{3}\right)^{6}
$$

which means that $D_{1}=\left(\frac{4}{3}\right)^{6}$.