6. Real numbers $a, b, c, d$ satisfy the equation $a b + b c + c d + d a = 16$.

a) Prove that among the numbers $a, b, c, d$, there are two with a sum of at most 4.

b) What is the minimum value that the sum $a^{2} + b^{2} + c^{2} + d^{2}$ can have?
SOLUTION. a) From the equality $16=a b+b c+c d+d a=(a+c)(b+d)$, it follows that both sums $a+c$ and $b+d$ cannot be greater than 4 at the same time. Therefore, at least one of the sums $a+c$ or $b+d$ must have the required property.

b) Let us use the general equality

$$
a^{2}+b^{2}+c^{2}+d^{2}=\frac{1}{2}(a-b)^{2}+\frac{1}{2}(b-c)^{2}+\frac{1}{2}(c-d)^{2}+\frac{1}{2}(d-a)^{2}+a b+b c+c d+d a
$$

whose validity can easily be verified by rearranging the right side. Given the non-negativity of the squares $(a-b)^{2},(b-c)^{2},(c-d)^{2}$, and $(d-a)^{2}$, we obtain the estimate for the left side of the equality

$$
a^{2}+b^{2}+c^{2}+d^{2} \geqq a b+b c+c d+d a=16 \text {. }
$$

Is the number 16 the smallest value of the considered sums? In other words: does equality occur in the derived inequality for some suitable quadruple? From our approach, it is clear that we need to decide whether for some of the considered quadruples, $a-b=b-c=c-d=d-a=0$ holds, i.e., $a=b=c=d$. For such a quadruple, the equation $a b+b c+c d+d a=16$ takes the form $4 a^{2}=16$, which is satisfied by $a= \pm 2$. For the suitable quadruples $a=b=c=d=2$ and $a=b=c=d=-2$, the sum $a^{2}+b^{2}+c^{2}+d^{2}$ indeed has the value 16, so it is the desired minimum.

GUIDING AND SUPPLEMENTARY PROBLEMS: