<s>[INST] Smallest value of $a^2 + b^2 + c^2+ d^2$, given values for $(a+b)(c+d)$, $(a+c)(b+d)$, and $(a+d)(b+c)$ 
If $a$, $b$, $c$, $d$ belong to $\mathbb{R}$, and
  $$(a+b)(c+d)=143 \qquad (a+c)(b+d)=150 \qquad (a+d)(b+c)=169$$
  Find the smallest possible value of 
  $$a^2 + b^2 + c^2+ d^2$$

I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.
Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.
 [/INST] Let $\lambda = a + b + c + d$. When you sum over the 3 equations.
$$(a+b)(c+d)=143,\quad (a+c)(b+d)=150,\quad (a+d)(b+c)=169\tag{*1}$$
LHS sums to
$$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
while RHS sums to $462$. This leads to
$$a^2 + b^2 + c^2 + d^2 = \lambda^2 - 462$$
To minimize $a^2+b^2+c^2+d^2$, one just need to minimize $\lambda^2$. If you look at LHS
of the 3 equations, all of them is a product of $2$ factors which sum to $\lambda$.
In general, if we have $p + q = \lambda$, then
$$4pq = (p+q)^2 - (p-q)^2 \le (p+q)^2 = \lambda^2$$
The set of 3 equations tell us
$$\begin{align}\lambda^2 &\ge 4\max\{ (a+b)(c+d), (a+c)(b+d), (a+d)(b+c) \}\\
&= 4\max\{ 143, 150, 169 \}\\&= 676\end{align}$$
As a result, 
$$a^2 + b^2 + c^2 + d^2 = \lambda^2 - 462 \ge 214$$
To see $214$ is the actual minimum, we need to find $(a,b,c,d)$ which satisfies $(*1)$ and $\lambda^2 = 676 = 26^2$.
Flipping all the signs of $a,b,c,d$ if necessary, we can assume $\lambda = 26$.
The third equation $(a+d)(b+c) = 169 = 13^2 = \frac14 (26)^2$ tell us
$a + d = b + c$.
Introduce $u,v$ such that
$$(a,b,c,d) = \left( \frac{13+u}{2}, \frac{13+v}{2}, \frac{13-v}{2}, \frac{13-u}{2}\right)$$
and substitute into the first and second equation and simplify, we obtain
$$\left(\frac{u+v}{2}\right)^2 = 26\quad\text{ and }\quad \left(\frac{u-v}{2}\right)^2 = 19$$
Using this, we find following 4-tuple
$$(a,b,c,d) = {\small \left(
\frac{13+\sqrt{26}+\sqrt{19}}{2},
\frac{13+\sqrt{26}-\sqrt{19}}{2},
\frac{13-\sqrt{26}+\sqrt{19}}{2},
\frac{13-\sqrt{26}-\sqrt{19}}{2}
\right)}$$
is a solution of the 3 equations in $(*1)$ with $a + b + c + d = 26$. 
One can verify $a^2 + b^2 + c^2 + d^2 = 214$ for this particular solution. As a result, the lower bound $214$ is achievable and $214$ is the smallest possible value we seek.
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