MHB What is the Minimum Value of Sum of Squares for Real Numbers in this POTW?

[anemone]

Here is this week's POTW:

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Let $a,\,b,\,c$ and $d$ be real numbers that satisfy

$(a+b)(c+d)=13,\\(a+c)(b+d)=24,\\(a+d)(b+c)=25.$

Find the smallest possible vale of $a^2+b^2+c^2+d^2$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the castor28 for his correct solution, which you can find below: (Smile)

Spoiler

We note that the given equations imply that $(a+b+c+d)^2 = a^2 + b^2 + c^2 + d^2 + 13 + 24 + 25$.

If we write $t=a+b+c+d$, $x=a+b$, $y=a+c$, and $z=a+d$, the equations can be written as:
$$\begin{align*}
x(t-x) &= 13\qquad[1]\\
y(t-y) &= 24\qquad[2]\\
z(t-z) &= 25\qquad[3]
\end{align*}
$$

As these quadratic equations must have real solutions, their discriminants must be non-negative. The strongest condition comes from the third equation and gives $t^2\ge100$. As changing the sign of $t$ merely changes the signs of all the variables, we may assume that $t\ge10$.

It remains to show that this value can be effectively attained. We may compute a solution explicitly. Taking $t = 10$ and choosing $y_1=6$ and $z_1=5$ as roots of the least two equations, we get the system:
$$\begin{align*}
a+c &=6\\
b+d &= 4\\
a+d&=5\\
b+c&=5
\end{align*}
$$

and the general solution of that system is $a=5-d$, $b=4-d$, $c=d+1$. Substituting in the first given equation and solving for $d$, we get $d=2\pm\sqrt3$. This gives one possible solution:
$$\begin{align*}
a &= 3- \sqrt3\\
b &= 2-\sqrt3\\
c &= 3+\sqrt3\\
d &= 2+\sqrt3
\end{align*}
$$