Find the minimum value of $T=a^2+b^2+c^2+d^2$ Given reals $a,b,c,d$ such that $$\left\{\begin{matrix}(a+b)(c+d)=2 
 &  & \\(a+c)(b+d)=3
 &  & \\ (a+d)(b+c)=4
 &  & 
\end{matrix}\right..$$
Find the minimum value of  $  T=a^2+b^2+c^2+d^2.$
I noticed that $(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=2ab+2ac+2ad+2bc+2bd+2cd\leq 3(a^2+b^2+c^2)=3T$ 
by applying the inequality $x^2+y^2\geq2xy$ $\forall x,y\in \mathbb R$. So $T\geq 3$ . However, equality doesn't occur with this method. 
Could you help me solve the problem?

The answer is 7.


By your work and by AM-GM we obtain:
$$a^2+b^2+c^2+d^2=(a+b+c+d)^2-2-3-4\geq4(a+d)(b+c)-9=7.$$
The equality occurs for example, for $a+d=b+c=2$, 
which gives that the equality indeed occurs:
$$d=2-a,$$ $$c=2-b,$$
$$(a+b)(4-a-b)=2$$ and
$$(a-b+2)(b-a+2)=3.$$
Can you end it now?
