Prove that $(ac+bd)^2 + (ad-bc)^2 \geq 144$ , if $a+b=4, c+d=6$. I got that $(ac)^{2}+(bd)^{2}+(ad)^{2}+(bc)^{2} \geq 144$. How is this one solved?

$$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$$
Method $\#1:$
Assuming $a,b$ are real, 
$(a-b)^2\ge0\iff a^2+b^2\ge2ab\iff2(a^2+b^2)\ge(a+b)^2=4^2$
Similarly, for $c^2+d^2$
Method $\#2:$
Assuming $a,c$ are real, 
$a^2+b^2=a^2+(4-a)^2=2(a^2-4a+8)=2[(a-2)^2+4]\ge2\cdot4$
Similarly, for $c^2+d^2$
See also: Brahmagupta-Fibonacci Identity
