easy with caushy: $ \left(\frac{c}{a+2b}+\frac{a}{b+2c}+\frac{b}{c+2a}\right)(3(ab+bc+ac))\geq (a+b+c)^2$ $ \Leftrightarrow \left(\frac{c}{a+2b}+\frac{a}{b+2c}+\frac{b}{c+2a}\right) \geq \frac{(a+b+c)^2}{3(ab+bc+ac)}\geq1$

Wasn't the same thing written by ghjk? Well one more method: Using Cauchy Lema We get the given expression $ \ge \dfrac{(\sum a)^2}{3\sum ab} \ge 1$ from the well known inequalities. Though this is the same, just posted it for those who wanted to see it done by this way.

Aravind Srinivas L wrote: Wasn't the same thing written by ghjk? Well one more method: Using Cauchy Lema We get the given expression $ \ge \dfrac{(\sum a)^2}{3\sum ab} \ge 1$ from the well known inequalities. Though this is the same, just posted it for those who wanted to see it done by this way. You post was also the same as ghjk I see no difference in Cauchy Schwartz and Cauchy Lemma.

Johan Gunardi wrote: For the positive real numbers $ a,b,c$ prove the inequality \[ \frac {c}{a + 2b} + \frac {a}{b + 2c} + \frac {b}{c + 2a}\ge1. \] Let $ x=a + 2b,y=b + 2c,x=c + 2a$ then we must prove $ \sum_{cyc} \frac{4y+z-2x}{9x}\ge 1$ or $ \sum_{cyc}\frac{4y+z}{x} \ge 15$ its obviously true by AM-GM

For $ a,\ b,\ c$, prove that $ \frac{2007c}{2008a+2009b}+\frac{2008a}{2009b+2007c}+\frac{2009b}{2007c+2008a}\geq \frac{3}{2}$.

kunny wrote: For $ a,\ b,\ c$, prove that $ \frac {2007c}{2008a + 2009b} + \frac {2008a}{2009b + 2007c} + \frac {2009b}{2007c + 2008a}\geq \frac {3}{2}$. Let $ 2007c=x$, $ 2008a=y$, and $ 2009b=z$, then our inequality takes form \[ \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge 1.5\]

a,b,c are positive real numbers ??? ... sorry I'm not good at English

@kfc9512: it should be positive real lha.. assume that it's not positive, then let $ x=0,y=-1,z=2$ then $ LHS=0-\frac{1}{2}-2<\frac{3}{2}$.

Sorry .... Also.. i'm not good at math.. thank you.. (and... \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge 1.5by caushy)-just practice

I've just found a really beautiful proof to this problem (inspired by a proof of Nesbitt's Inequality): Let $ P=\sum\frac{c}{a+2b}$, $ Q=\sum\frac{a}{a+2b}$, $ R=\sum\frac{b}{a+2b}$. By AM-GM, we have $ P+2Q=\sum\frac{c+2a}{a+2b}\ge3$ and $ 2P+R\ge\frac{2c+b}{a+2b}\ge3$. So $ 9P+2Q+4R=4(2P+R)+(P+2Q)\ge15$, but $ 2Q+4R=2(Q+2R)=6$, so $ P\ge1$. Q.E.D.

Another approach is by brute force, after computations we get $ 2\sum_{cyc}{a^3} \geq 3abc + \sum_{cyc}{ a^2 (3c - 2b)} \Leftrightarrow {\underbrace{\sum_{cyc}{a^3} - 3abc}}_{\geq 0 } + {\underbrace{\sum_{cyc}{a^2(a + 3c -2b)}}_{\geq 0} \geq 0}$ for second sum we must prove somehow that is greater than zero

Johan Gunardi wrote: For the positive real numbers $ a,b,c$ prove the inequality \[ \frac {c}{a + 2b} + \frac {a}{b + 2c} + \frac {b}{c + 2a}\ge1. \] $ \sum\frac{a}{b+2c}=\frac{a^2}{ab+2ca}\geq \frac{(a+b+c)^2}{3\sum ab}\geq 1$ from the trivial ineq

See #2 popst. You should see the former post before you post.

Hi Use titu"s lemma.

it is just Cauchy Schwartz

am i watching the similar problem too many times? I saw it above which the difference is just the number of variant became 4 instead of 3

$c^2/(ac+2bc)+a^2/(ab+2ac)+b^2/(bc+2ab)>=1$ $(a+b+c)^2/3(ab+ac+bc)>=1$ $a^2+b^2+c^2>=ab+ac+bc$

Titu lemma and easy problem

$\frac{c^2}{ac+2bc}+\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}\geq\frac{(a+b+c)^2}{3(ab+ac+bc)}\geq1$

My Approach using Titu's lemma \[ \frac{a}{b + 2c} + \frac{b}{c + 2a} + \frac{c}{a + 2b} = \frac{a^{2}}{a(b + 2c)} + \frac{b^{2}}{b(c + 2a)} + \frac{c^{2}}{c(a + 2b)} \]\[ \frac{a^{2}}{a(b + 2c)} + \frac{b^{2}}{b(c + 2a)} + \frac{c^{2}}{c(a + 2b)} \geq \frac{(a + b + c)^{2}}{\sum_{\text{cyc}} a(b + 2c)} \geq 1 \]\[ \implies \frac{a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)}{3(ab + bc + ca)} \geq 1 \] \[ a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) \geq 3(ab + bc + ca) \]\[ a^{2} + b^{2} + c^{2} \geq ab + bc + ca \]$$ \blacksquare $$

jgnr wrote: For the positive real numbers $ a,b,c$ prove the inequality \[ \frac {c}{a + 2b} + \frac {a}{b + 2c} + \frac {b}{c + 2a}\ge1. \] Hey Bro use Titu's Lemma like we can write it as $ac/(a^2+2ab)+ab/(b^2+2bc)+bc/(c^2+2ac)>=a^2/(a^2+2ab)+b^2/(b^2+2bc)+c^2/(c^2+2ac)$ then after that we use titu's lemma: $>=(a+b+c)^2/(a+b+c)^2>=1$

Let $ a,b,c >0 . $ Prove that $$ \frac {a}{b + kc} + \frac { b}{c +ka} + \frac {c}{a + kb} \geq \frac {3}{k + 1}$$Where $ k>0. $

sqing wrote: Let $ a,b,c >0 . $ Prove that $$ \frac {a}{b + kc} + \frac { b}{c +ka} + \frac {c}{a + kb} \geq \frac {3}{k + 1}$$Where $ k>0. $ I did using Titu's Lemma, \[ \frac {a}{b + kc} + \frac { b}{c +ka} + \frac {c}{a + kb} = \frac {a^{2}}{a(b + kc)} + \frac { b^{2}}{b(c +ka)} + \frac {c^{2}}{c(a + kb)} \] Claim : $\frac {a^{2}}{a(b + kc)} + \frac { b^{2}}{b(c +ka)} + \frac {c^{2}}{c(a + kb)} \geq \frac{(a + b + c)^{2}}{(ab + bc + ca)(k + 1)} \geq \frac {3}{k + 1}$ Proof : $\frac{(a + b + c)^{2}}{(ab + bc + ca)(k + 1)} \geq \frac {3}{k + 1} = \frac{(a + b + c)^{2}}{(ab + bc + ca)} \geq 3 $ $(a + b + c)^{2} \geq 3(ab + bc + ca) \implies a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) \geq 3(ab + bc + ca)$ $ \implies a^{2} + b^{2} + c^{2} \geq ab + bc + ca$ $ \blacksquare $

Rmo level problem $\frac{c^2}{ac + 2b} + \frac{a^2}{ab + 2ac} + \frac{b^2}{bc + 2ab} \geq \frac{(a + b + c)^2}{3(ab + ac + bc)}$ by Titu's, and $(a + b + c)^2 \geq 3abc + 3ac + 3bc$ since $a^2 + b^2 + c^2 \geq ab + ac + bc$, so we are done.