wangzishan wrote: Leta,b,c are postive real numbers,proof that $ \frac {a}{b + 2c} + \frac {b}{c + 2a} + \frac {c}{a + 2b}\geq1$ Just standard CS: $ \left ( \sum_{cyc} a(b+2c) \right ) \left ( \sum_{cyc} \frac{a}{b+2c} \right ) \ge \left ( \sum_{cyc} a \right )^2$. So it suffices to prove: $ \left ( \sum_{cyc} a \right )^2 \ge \left ( \sum_{cyc} a(b+2c) \right ) \iff$ $ a^2+b^2+c^2 \ge ab+bc+ca$, and we are done.

Mathias_DK wrote: wangzishan wrote: Leta,b,c are postive real numbers,proof that $ \frac {a}{b + 2c} + \frac {b}{c + 2a} + \frac {c}{a + 2b}\geq1$ Just standard CS: $ \left ( \sum_{cyc} a(b + 2c) \right ) \left ( \sum_{cyc} \frac {a}{b + 2c} \right ) \ge \left ( \sum_{cyc} a \right )^2$. So it suffices to prove: $ \left ( \sum_{cyc} a \right )^2 \ge \left ( \sum_{cyc} a(b + 2c) \right ) \iff$ $ a^2 + b^2 + c^2 \ge ab + bc + ca$, and we are done. nice work

Here is another approach: $ \sum \frac {a}{b+2c}= \sum \frac {a^2}{ab+2ac} \ge \sum \frac {(a+b+c)^2}{3(ab+bc+ca)}$. But it is easy to see that: $ a^2+b^2+c^2 \ge 3(ab+bc+ca)$. Our proof is completed

This inequality is from Czech and Slovak match 1999 See here

By the Cauchy-Schwarz Inequality, we know that: $$\left( \frac{a}{b + 2c} + \frac{b}{c + 2a} + \frac{c}{a + 2b} \right) \left(a(b+2c)+b(c+2a)+c(a+2b) \right) \ge (a+b+c)^2 \Longleftrightarrow \frac{a}{b + 2c} + \frac{b}{c + 2a} + \frac{c}{a + 2b} \ge \frac{(a+b+c)^2}{3ab+3bc+3ca}$$Since $$a^2+b^2+c^2 \ge ab + bc + ca \Longleftrightarrow (a+b+c)^2 \ge 3(a+b+c) \Longleftrightarrow \frac{(a+b+c)^2}{a+b+c} \ge 3$$it follows that $$\frac{a}{b + 2c} + \frac{b}{c + 2a} + \frac{c}{a + 2b} \ge \frac{(a+b+c)^2}{3ab+3bc+3ca} \ge 1$$as desired.