It is so easy, even i can prove it. $3xyz(x+y+z)\le (xy+yz+zx)^2\le (x^2+y^2+z^2)^2=1$

0≤(a-b)^2+(b-c)^2+(c-a)^2 therefore ab+bc+ca ≤ a^2+b^2+c^2.hence 3(ab+bc+ca) ≤ (a+b+c)^2 if a=xy , b=yz , c=zx 3xyz(x+y+z) ≤ (xy+yz+zx)^2 ≤ (x^2+y^2+z^2)^2 = 1

To #2: 　　Sorry but I am not quite familiar with the inequality you use. Would you mind explaining how to get '$ 3xyz(x+y+z)\le (xy+yz+zx)^{2}$'? 　　I get it by: 　　　　$3(1/xy+1/yz+1/zx)\le (1/x+1/y+1/z)^2$ 　　　　$ \Leftarrow 1/xy+1/yz+1/zx\le 1/x^2+1/y^2+1/z^2$ To #3: 　　Sorry but I am puzzled with the 'hence' you use at line3 because $3(a^2+b^2+c^2)\ge(a+b+c)^2$.

Hi! tuarek just added $2(ab+bc+ca)$ to the both sides of his inequality,i think the "hence" is clear now!

ArrowRowe wrote: To #2: 　　Sorry but I am not quite familiar with the inequality you use. Would you mind explaining how to get '$ 3xyz(x+y+z)\le (xy+yz+zx)^{2}$'?. that's a known inequality. by substituting $xy=a, yz=b, zx=c$, you can easily show the inequality is equivalent to $3(ab+bc+ca)\le (a+b+c)^2$ which is easy

Oh please forgive me... Thanks a lot for your patience!

$ 3(x^{2}+y^{2}+z^{2})>=(x+y+z)^{2}) $ $ (x+y+z)^{3}>=27(xyz)$