<s>[INST] Maximizing $a^2+b^2+c^2+d^2$ with given constraints The following problem is from a local contest which ended today:

Let $a,b,c,d$ be positive real numbers such that $$(a+b)(c+d)=143\\ (a+c)(b+d)=150\\ (a+d)(b+c)=169$$ Find the maximum value of $a^2+b^2+c^2+d^2$.

Here are my workings:
We have $$\tag{1}ac+bc+bd+ad=143$$ $$\tag{2}{ab+bc+cd+ad=150}$$ $$\tag{3}{ab+bd+ac+cd=169}$$
Summing $(1)$, $(2)$ and $(3)$, we have $$2(ab+bc+cd+ac+ad+bd)=462$$
We have $$\begin{align} a^2+b^2+c^2+d^2 &=(a+b+c+d)^2-2(ab+bc+cd+ac+ad+bd)\\ &= (a+b+c+d)^2-462\end{align}$$
So we have to maximize $a+b+c+d$.
I can't proceed from here. Minimizing the expression seems easy by Cauchy-Schwarz or AM-GM but I don't know how to maximize this.
 [/INST] As @CalvinLin pointed out,
under the conditions, $a^2 + b^2 + c^2 + d^2$ is not bounded from above.
Explanation:
Let $t\in (0, 1)$ and
$$a = \frac{t}{2} + \frac{Q}{4t},
\quad b = -\frac{t}{2} + \frac{Q}{4t},
\quad c = \frac{143t}{Q} + \frac{19}{2t},
\quad d = \frac{143t}{Q} - \frac{19}{2t}$$
where $Q = \sqrt{ {t}^{4}+638\,{t}^{2}+361} - \sqrt{{t}^{4}+66\,{t}^{2}+361}$.
It is not difficult to verify
that
$a, b, c, d > 0$, and
$(a + b)(c + d) = 143$, and $(a + c)(b + d) = 150$,
and $(a + d)(b + c) = 169$.
However, $a^2 + b^2 + c^2 + d^2 = t^2 + 176 + \frac{361}{t^2} \to \infty$ as $t \to 0^{+}$.

Remarks: Explain how to obtain the above $a, b, c, d$.
Consider the system of equations
\begin{align*}
 (a + b)(c + d) = 143, \tag{1}\\
 (a + c)(b + d) = 150, \tag{2}\\
 (a + d)(b + c) = 169. \tag{3}
\end{align*}
(1) gives
$d = \frac{143}{a + b} - c$.
[(2) - (3)] gives
$(b-a)c + ad - db + 19 = 0$
which results in $c = \frac{81a - 62b}{a^2 - b^2}$.
We see that if $a > b$ and $a-b \to 0^{+}$, then $c \to \infty$. So, we let $a - b = t$.
Then we have
$$4\,{t}^{2}{b}^{4}+8\,{t}^{3}{b}^{3}+ \left( 5\,{t}^{4}-352\,{t}^{2}-
361 \right) {b}^{2}+ \left( {t}^{5}-352\,{t}^{3}-361\,t \right) b-88\,
{t}^{4}+5022\,{t}^{2}
 = 0.$$
We can solve $b$ in closed form. Indeed, dividing both sides by $4t^2$,
we have
$${b}^{4}+2\,t{b}^{3}+ \left( -88+ \frac{5}{4}t^2 -{\frac {361}{4t^2}} \right) {b}^{2}+ \left( -{\frac {361}{4t}} + \frac14{t}^{3}-88
\,t \right) b+{\frac {2511}{2}}-22\,{t}^{2} = 0.
$$
We eliminate the term $b^3$ by letting $b = z - t/2$:
$${z}^{4}+ \left( -{\frac {361}{4t^2}}-88-\frac14{t}^{2} \right) {z
}^{2}+{\frac {20449}{16}} = 0.$$
Luckily, it can be solved in closed form.
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