Prove the inequality stated below: 
If $a, b, c, d > 0$ then prove:
$$\frac{a}{b+c+d} + \frac{b+c}{a+d} + \frac{d}{b+a} > 1$$

I tried to consider the difference and lead to a common denominator, but in the numerator, there remains ${1}$ term with a minus which is less than ${0}$

$a^3 - a^2 c + b (b + c)^2 + c d^2 + d^3 + a (b + c) (c + d)$


Notice that if $m,n,k>0$ then $${m\over n} >{m\over n+k}$$ so
$$\frac{a}{b+c+d} + \frac{b+c}{a+d} + \frac{d}{b+a} >\frac{a}{\color{red}{a}+b+c+d} + \frac{b+c}{\color{red}{b+c}+a+d} + \frac{d}{\color{red}{c+d}+b+a}$$ $$= \frac{a+b+c+d}{a+b+c+d}=1$$ 
