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.

Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$
$$2ac+2ad+2bc+2bd+2ab+2cd \leq 3a^2+3b^2+3c^2+3d^2$$
So $$a^2+b^2+c^2+d^2\geq 154$$
But, unfortenuly this is not the minumum, since variables can not be all the same.
