Solve Maximum of a+b+c+d Without Wolfram

[Numeriprimi]

Hey!
I found one difficult equation for me...
It is:
Valid for real numbers a, b, c, d:
a + b = c + d
ad = bc
ac + bd = 1
What is the maximum of a+b+c+d?

And DON'T SAY wolfram, really no wolfram... I used it and it isn't good.
So... have you got any idea?

 

[Michael Redei]

Add $a$c to both sides of the second equation, and you get
$$
ad + ac = ac + bc
\iff
a(c+d) = (a+b)c = (c+d)c.
$$
One possibility now would be $c+d=0$, but then you'd also have $a+b=0$, and so $a+b+c+d=0$. So suppose $c+d\neq0$, then $a=c$, and from the first equation you also have $b=d$.

Now you're really left with only two unknowns, which you can use in the third equation to make $a+b+c+d=2a+2b$ minimal.

 

[Numeriprimi]

Hmm... Sorry, bud I don't understand last line. Ok, a+b+c+d=2a+2b, yes, but how I find numerical value of a+b+c+d in second variant?

 

[Michael Redei]

Your third equation says $ac+bd=1$, i.e. $a^2+b^2=1$, so if you assume $b\geq0$ (since you want $2a+2b$ to be maximal), you get $b=\sqrt{1-a^2}$. This means you need to find a value $a$ for which $2a+2b = 2a+2\sqrt{1-a^2}$ is maximal.

Do you know how to find such an $a$?