1 College math competitions ranking

Question

Let $a,b,c,d$ be real numbers which are strictly positive.
Show that $$1\lt\dfrac{a}{a+b+c}+\dfrac{b}{b+a+d}+\dfrac{c}{c+a+d}+\dfrac{d}{d+c+b}\lt2.$$ I need help please.

Answer 1

No advanced techniques are necessary, just observe that $$ \frac{a}{a+b+c+d}<\frac{a}{a+b+c}<\frac{a}{a+b}. $$

Answer 2

Because $1=\sum\limits_{cyc}\frac{a}{a+b+c+d}<\sum\limits_{cyc}\frac{a}{a+b+c}<\sum\limits_{cyc}\frac{a+d}{a+b+c+d}=2$