Are these numbers real?

When $x \in \mathbb{R}^+$ , $x+\dfrac{1}{x} \geq 2$ .

$\dfrac{a+b+c}{d} + \dfrac{b+c+d}{a} + \dfrac{c+d+a}{b} + \dfrac{d+a+b}{c} \\ = \dfrac{a}{d}+\dfrac{b}{d}+\dfrac{c}{d}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{d}{a}+\dfrac{c}{b}+\dfrac{d}{b}+\dfrac{a}{b}+\dfrac{d}{c}+\dfrac{a}{c}+\dfrac{b}{c} \\ = \dfrac{a}{d}+\dfrac{d}{a}+\dfrac{b}{d}+\dfrac{d}{b}+\dfrac{c}{d}+\dfrac{d}{c}+\dfrac{b}{a}+\dfrac{a}{b}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{c} \\ \geq 2+2+2+2+2+2 \\ \geq \boxed{12}$

$\frac { a+b+c }{ d } +\frac { b+c+d }{ a } \frac { a+c+d }{ b } +\frac { a+b+d }{ c } =\frac { a+b+c }{ d } +1+\frac { b+c+d }{ a } +1\frac { a+c+d }{ b } +1+\frac { a+b+d }{ c } +1-4\\ =\frac { a+b+c+d }{ d } +\frac { a+b+c+d }{ a } \frac { a+b+c+d }{ b } +\frac { a+b+c+d }{ c } -4=(a+b+c+d)\left( \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } \right) -4$

from cauchy schwarz ineq we have $(a+b+c+d)\left( \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } \right) \ge { \left( 1+1+1+1 \right) }^{ 2 }=16$

so $(a+b+c+d)\left( \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } \right) -4\ge { 12 }$

the minimum value is $12$

I just thought it simple. If the whole numbers a,b,c,d are 1, that's the minimum value in this case.

(this is only a trick for guessing and is not a proper solution)

For a symmetrical expression, the max or min usually happens when all the terms are the same. Changing everything to x gives 12.

Here, i see the whole expression is symmetric along a,b,c,d. So, I will put all these numbers equall to each other to obtain the critical value ;) , i.e. a=b=c=d , which makes the expression equals to 12.