First of all, see easily that $\displaystyle\sum_{cyc} \dfrac{b}{2a+b}=\displaystyle\sum_{cyc} \dfrac{b^2}{2ab+b^2}\ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1$. Then we have $\displaystyle\sum_{cyc}\dfrac{a}{2a+b}=\dfrac{1}{2}\displaystyle\sum_{cyc}\dfrac{2a}{2a+b}=\dfrac{1}{2}\displaystyle\sum_{cyc}\left(1-\dfrac{b}{2a+b}\right)\le \dfrac{1}{2}(3-1)=1$.

Czech and Slovak1999: For arbitrary positive real numbers $ a,b,c$ prove the inequality\[ \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge 1.\]

sqing wrote: Czech and Slovak1999: For arbitrary positive real numbers $ a,b,c$ prove the inequality\[ \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge 1.\] $\sum_{cyc}\frac{a}{b+2c}\geq 1$ $\iff$ $\sum_{cyc}\frac{a^2}{ab+2ca}\geq 1$ By Titu, $\sum_{cyc}\frac{a^2}{ab+2ca}\geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$ By Rearrangement, This is $\geq 1$