An inequality: $ab=cd=1$, $(a,b,c,d \geq 0)$, prove that $(a+b)(c+d)+4 \geq 2(a+b+c+d)$

Question

$ab=cd=1$ $(a,b,c,d \geq 0)$, prove that $(a+b)(c+d)+4 \geq 2(a+b+c+d)$

Attempt 1:

$(a+b)(c+d)+4 = ac+ad+bc+bd+ab+cd+2=(ab+ac+bd+cd)+ad+bc+2=(a+d)(b+c)+ad+bc+2$

I was stuck here, so I have a new approach:

$ab=cd=1 \Leftrightarrow b= \frac{1}{a}, d= \frac {1}{a}$

$\Rightarrow (a+b)(c+d)+4 \geq 2(a+b+c+d) \Leftrightarrow \left(a+\frac{1}{a}\right)\left(c+\frac{1}{c}\right)+4 \geq 2\left(a+\frac{1}{a}+c+\frac{1}{c}\right)$

And I was stuck again...

Answer 1

If $x, y \ge 2$ then $xy -2(x+y)+4 =(x-2)(y-2) \ge 0$.

Apply this to $x = a+1/a, y = c+1/c$ since $a+1/a-2 =(\sqrt{a}-1/\sqrt{a})^2 \ge 0$.

Answer 2

Because by AM-GM $$(a+b)(c+d)+4-2(a+b+c+d)=(a+b-2)(c+d-2)\geq$$ $$\geq(2\sqrt{ab}-2)(2\sqrt{cd}-2)=0.$$