[nooonuii]

12. Czech and Slovak 2005 $a,b,c>0, abc =1$
$$\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq \frac{3}{4}$$

Solution :

Brute force!! The inequality is equivalent to
$$ab+bc+ca+a+b+c\geq 6$$
which is easy to prove by AM-GM inequality.

13. Japan 2005 $a,b,c>0, a+b+c =1$
$$a\sqrt[3]{1+b-c}+b\sqrt[3]{1+c-a}+c\sqrt[3]{1+a-b}\leq 1$$

Solution :

$\displaystyle{LHS=a\sqrt[3]{(a+2b)\cdot 1\cdot 1}+b\sqrt[3]{(b+2c)\cdot 1\cdot 1}+c\sqrt[3]{(c+2a)\cdot 1\cdot 1}}$

$\displaystyle{\leq \frac{a(a+2b+2)+b(b+2c+2)+c(c+2a+2)}{3}}$

$\displaystyle{=\frac{(a+b+c)^2+2(a+b+c)}{3}}$

$=1$