5. For positive real numbers $a, b, c, d$ the following holds:

$$
a+b=c+d, \quad a d=b c, \quad a c+b d=1
$$

What is the maximum value that the sum $a+b+c+d$ can have?
SOLUTION. First, we will show that the first two equalities from the problem statement are satisfied only when $a=c$ and simultaneously $b=d$. Indeed, due to the given numbers being positive (and thus different from zero), we can rewrite the mentioned equalities as

$$
a\left(1+\frac{b}{a}\right)=c\left(1+\frac{d}{c}\right) \quad \text { and } \quad \frac{b}{a}=\frac{d}{c}.
$$

From the second equality, we see that the sums inside the parentheses on both sides of the first equality have the same positive value, so the first factors on both sides of the equation must be equal. Therefore, $a=c$, from which it follows that $b=d$.

Knowing that $a=c$ and $b=d$, we can further work only with the variables $a$ and $b$ and find the maximum value of the given sum

$$
S=a+b+c+d=2(a+b)
$$

under the sole condition that the positive numbers $a, b$ satisfy the equality $a^{2}+b^{2}=1$, which is the expression of the third given equality $a c+b d=1$ (the first two are evident due to the equalities $a=c$ and $b=d$).

Notice that for the square (of the positive) sum $S$ we have

$$
S^{2}=4(a+b)^{2}=4\left(a^{2}+b^{2}\right)+8 a b=4 \cdot 1+8 a b=4(1+2 a b),
$$

so the value of $S$ will be the greatest precisely when the value of $2 a b$ is the greatest. From the obvious inequality $(a-b)^{2} \geqq 0$ after expanding, we get

$$
2 a b \leqq a^{2}+b^{2}=1
$$

with equality $2 a b=1$ occurring precisely when $a=b$, which for positive numbers $a, b$ together with the condition $a^{2}+b^{2}=1$ leads to the only suitable pair $a=b=1 / \sqrt{2}$. The maximum value of the expression $2 a b$ is thus 1, so the maximum value of the expression $S^{2}$ is $4(1+1)=8$, and therefore the maximum value of $S$ is $\sqrt{8}=2 \sqrt{2}$. It is achieved for the only admissible quadruple $a=b=c=d=1 / \sqrt{2}$.

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