Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$ 
Prove that
  $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
  with $a>0, b>0 , c> 0$ and $d>0.$

My attempt:
$$\begin{align*}\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)& =  \dfrac{abcd+b^2c^2+a^2d^2+abcd}{abcd}\\
    & =\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
     &=\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{(ad)^2+(bc)^2+2(ad)(bc)}{abcd}\\
&=\dfrac{(ad+bc)^2}{abcd}\end{align*}$$
I don't know how to continue from this.
Can someone help me?
 A: Use AM-GM. $\frac{ad + bc}{2} \ge \sqrt{abcd}$. Squaring both sides, you get the answer. A tiny tip: If everything is positive, and you have an inequality, think about AM-GM once at least. 
A: Taking a different approach entirely, note that
$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)=1+{ad\over bc}+{bc\over ad}+1$$
Thus, letting $ad/bc=x$ and noting that $x\gt0$, we see that
$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)\ge4\iff x+{1\over x}\ge2\iff x^2-2x+1\ge0\iff(x-1)^2\ge0$$
(Note, the stipulation $x\gt0$ is important is multiplying both sides of the inequality $x+1/x\ge2$ by $x$ to get to $x^2+1\ge2x$.)
A: Now,     $$\frac{(ad+bc)^2}{abcd}-4=\frac{a^2d^2-2abcd+b^2c^2}{abcd}=\frac{(ad-bc)^2}{abcd}\geq0.$$Also, by C-S
$$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq\left(\sqrt{\frac{b}{a}\cdot\frac{a}{b}}+\sqrt{\frac{d}{c}\cdot\frac{c}{d}}\right)^2`=4$$
A: To continue from
$\tag 1 \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) = \dfrac{(ad+bc)^2}{abcd}$
set 
$\quad u = ad$
and
$\quad v = bc$
Then substituting in the rhs of $\text{(1)}$, we have
$\quad \dfrac{(u+v)^2}{uv} \ge 4 \text{ iff } (u-v)^2 \ge 0$
Note that if we let $a,b,c,d \in \Bbb R$ satisfy $abcd \gt 0$ then 
$\quad  \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) \ge 4$
and if $abcd \lt 0$ then
$\quad  \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) \le 4$
A: Let $x=a/b$, $y=c/d$, you'll get $2+x/y+y/x$.  Now use that for any positive number the sum of that number and its reciprocal is at least $2$.
