Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$ Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$
My attempt:I used two ways but I get to a wrong answer.
My first way:We know that $\frac{a}{b}+\frac{b}{a} \ge 2$ where $a,b \ge 0$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}=$
$\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \ge 4$
$+$
$\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\ge 4\sqrt{\frac{b}{a}*\frac{c}{b}*\frac{d}{c}*\frac{a}{d}}=4$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}\ge 8$
Then we will get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 0$
Which is not true.
My second way:I don't have enough time then I just explain it.
I used caushy-shuartz and I get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 2$
 A: $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac} \ge \frac{(a+b+c+d)^2}{ab+bc+cd+da+2ac+2bd}=\frac{(a+b+c+d)^2}{(a+c)(b+d)+2ac+2bd}$$
From Cauchy. 
$$(a+b+c+d)^2=(a+c)^2+(b+d)^2+2(a+c)(b+d) \ge 4ac +4bd+2(a+c)(b+d)$$
Since $(x+y)^2 \ge 4xy$.
So $$\sum_{cyc}\frac{a}{b+c} \ge 2 >\frac{1}{2}$$
A: $\sum\limits_{cyc}\frac{a}{b+c}=\sum\limits_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+s)^2}{\sum\limits_{cyc}(ab+ac)}\geq2$
Because the last inequality it's $(a-c)^2+(b-d)^2\geq0$. Done!
A: Let 
$$P=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}$$
$$Q=\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}+\frac{a}{a+b}$$
$$R=\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}+\frac{b}{a+b}$$
We have $$Q+R=4\tag{1}$$
$$P+Q=\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}+\frac{d+a}{a+b} \overbrace{\geq}^{\color{red}{\text{AM} \geq \text{GM}}} 4\tag{2}$$
$$\begin{align} P+R=\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\\ = \left(\frac{a+c}{b+c}+\frac{a+c}{d+a}\right)+ \left(\frac{b+d}{c+d}+\frac{b+d}{a+b}\right) \\
\overbrace{\geq}^{\color{blue}{\text{Titu's Lemma}}} \frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}=4\tag{3}
\end{align}
$$
Using $(1),(2)$ and $(3)$, we get the desired result.
