
4. Among real numbers $a, b, c$, and $d$ which satisfy

$$
a b+c d=a c+b d=4 \quad a \quad a d+b c=5
$$

find these, for which the value of $a+b+c+d$ is the least possible. Find this (the least) value as well.

(Jaromír Šimša)


Solution. We have

$$
\begin{aligned}
(a+b+c+d)^{2} & =a^{2}+b^{2}+c^{2}+d^{2}+2(a b+c d+a c+b d+a d+b c)= \\
& =a^{2}+b^{2}+c^{2}+d^{2}+2(4+4+5)=a^{2}+d^{2}+b^{2}+c^{2}+26 .
\end{aligned}
$$

Now $a^{2}+d^{2} \geqslant 2 a d, b^{2}+c^{2} \geqslant 2 b c$ where the equality holds iff $a=d$ and $b=c$ and from (1) we get

$$
(a+b+c+d)^{2} \geqslant 2 a d+2 b c+26=2 \cdot 5+26=36 .
$$

Thus among reals which satisfy the conditions we always have $a+b+c+d \geqslant 6$, where the equality holds iff $a=d$ and $b=c$, or

$$
2 a b=4, \quad a^{2}+b^{2}=5,
$$

thus $\{a, b\}=\{1,2\}$.

Conclusion. The sought $a, b, c, d$ are the quadruples $(1,2,2,1)$ and $(2,1,1,2)$ with the least value of $a+b+c+d=6$.

Final Round of the 61st Czech and Slovak

Mathematical Olympiad

(March 26-27, 2012)

$\mathbb{N} / /$ (6)
