it is Veitnam TST 2005 ,isn't it. the inequality equivalents to $\frac{1}{(1+x)^3}+\frac{1}{(1+y)^3}+\frac{1}{(1+z)^3} \ge \frac{3}{8}$ for $x=\frac{a}{b}$,$y=\frac{b}{c}$,$z=\frac{c}{a}$ so $xyz=1$ then prove this one http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112435#p112435 from china 2005 substitute $a=x,b=y,c=z,d=1$ and use power mean inequality

LightLucifer wrote: it is Veitnam TST 2005 ,isn't it. the inequality equivalents to $\frac{1}{(1+x)^3}+\frac{1}{(1+y)^3}+\frac{1}{(1+z)^3} \ge \frac{3}{8}$ for $x=\frac{a}{b}$,$y=\frac{b}{c}$,$z=\frac{c}{a}$ so $xyz=1$ then prove this one http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112435#p112435 from china 2005 substitute $a=x,b=y,c=z,d=1$ and use power mean inequality but here is $ \frac{1}{(1+x)^{3}} $ not $ \frac{1}{(1+x)^{2}} $ ,isn't it?

pxchg1200 wrote: LightLucifer wrote: it is Veitnam TST 2005 ,isn't it. the inequality equivalents to $\frac{1}{(1+x)^3}+\frac{1}{(1+y)^3}+\frac{1}{(1+z)^3} \ge \frac{3}{8}$ for $x=\frac{a}{b}$,$y=\frac{b}{c}$,$z=\frac{c}{a}$ so $xyz=1$ then prove this one http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112435#p112435 from china 2005 substitute $a=x,b=y,c=z,d=1$ and use power mean inequality but here is $ \frac{1}{(1+x)^{3}} $ not $ \frac{1}{(1+x)^{2}} $ ,isn't it? use power mean inequality then you will get a result

pxchg1200 wrote: Let $ a,b,c>0$ prove that:\[ \frac{a^{3}}{(a+b)^{3}}+\frac{b^{3}}{(b+c)^{3}}+\frac{c^{3}}{(c+a)^{3}}\geq \frac{3}{8} \]Good luck! See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=101425