<s>[INST] 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?
 [/INST] 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 $x^2+1\ge2x$.)
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