See here

I have seen that. And I need a legible post to add to the contests section. That one's LaTeX is absent. Also I look for other solutions than the one given there. I myself have a different one. I want to see others.

The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\ge \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\geq bc+ca+ab. \]

sqing wrote: The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\ge \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\geq bc+ca+ab. \] if $x = 2, y = 2, z = 1/142$ ${\it ssgm} \left( {\frac {x}{1+y+z}} \right) -{\it ssgm} \left( { \frac {yz}{1+y+z}} \right) ={\frac {158331}{303170}}$ But ${\it ssgm} \left( x \right) -{\it ssgm} \left( yz \right) =-{\frac {3} {142}} $

sqing wrote: The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\ge \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\geq bc+ca+ab. \] sqing 老师打印错了： The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\le bc+ca+ab. \]

xzlbq wrote: sqing wrote: The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\ge \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\geq bc+ca+ab. \] The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\le bc+ca+ab.---(1) \] arqady ot mudok,you must pay more working, because it is nice! (1)<=> Let $-27\,{u}^{3}-18\,{u}^{2}+ \left( -3+3\,{w}^{3}+27\,{v}^{2} \right) u+9 \,{v}^{2} \left( 1+{v}^{2} \right) \geq 0 $,prove that \[v^2\geq u\] Note: \[u=1/3\,x+1/3\,y+1/3\,z,v=1/3\,\sqrt {3\,xy+3\,zx+3\,yz},w=\sqrt [3]{x yz}\]

xzlbq wrote: arqady ot mudok,you must pay more working, add Luo Fang Xiang! If your 3 persons can do?

xzlbq wrote: xzlbq wrote: arqady ot mudok,you must pay more working, add Luo Fang Xiang! If your 3 persons can do? and this: jeff10 wrote: Supercali wrote: Let $a,b,c > 0$ such that $a+b+c \geq ab+bc+ca$. Prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{a+b+c}{2}$$ xzlbq wrote: More stronger: Let $a,b,c > 0$ such that $a+b+c \geq ab+bc+ca$. Prove that \[{\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}+\frac{3}{2}\,{\frac {ab+bc+ ac}{a+b+c}}\geq \frac{1}{2}\sum{a}+\frac{3}{2}\] 《=》 if $u\geq v^2$,prove \[18\,{u}^{4}-9\,{v}^{2}{u}^{3}+ \left( {w}^{3}-21\,{v}^{2} \right) {u}^ {2}+ \left( 9\,{v}^{4}+3\,{w}^{3} \right) u-{v}^{2}{w}^{3}\geq 0\]

xzlbq wrote: xzlbq wrote: arqady ot mudok,you must pay more working, add Luo Fang Xiang! If your 3 persons can do? 反证法即可（Can be reduced to absurdity）

xzlbq wrote: sqing wrote: The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\ge \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\geq bc+ca+ab. \] sqing 老师打印错了： The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\le bc+ca+ab. \]

xzlbq wrote: The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\le bc+ca+ab. \] We have $ A=\sum_{cyc} \frac{a-bc}{1+b+c} \le 0$. By Chebyshev: $\frac{(a+b+c-ab-bc-ca)}{3} \sum_{cyc} \frac{1}{1+b+c}\le A \le 0$. So, $a+b+c-ab-bc-ca\le 0$

mudok wrote: xzlbq wrote: The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\le bc+ca+ab. \] We have $ A=\sum_{cyc} \frac{a-bc}{1+b+c} \le 0$. By Chebyshev: $\frac{(a+b+c-ab-bc-ca)}{3} \sum_{cyc} \frac{1}{1+b+c}\le A \le 0$. So, $a+b+c-ab-bc-ca\le 0$ Well done!

mudok wrote: xzlbq wrote: The following inequality is also true. Let $a,b,c$ be positive reals satisfying $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\le \frac{bc}{1+b+c}+\frac{ca}{1+c+a}+\frac{ab}{1+a+b}.$Prove that :\[ a+b+c\le bc+ca+ab. \] We have $ A=\sum_{cyc} \frac{a-bc}{1+b+c} \le 0$. By Chebyshev: $\frac{(a+b+c-ab-bc-ca)}{3} \sum_{cyc} \frac{1}{1+b+c}\le A \le 0$. So, $a+b+c-ab-bc-ca\le 0$ we also have two other proofs

joybangla wrote: Let $a,b,c$ be positive reals satisfying : \[ \frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\ge \frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a} \]Then prove that : \[ \frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) \] Proposed by Dimitar Trenevski any proofs for this ugly main claim?