[nooonuii]

3. Romania 2005 $a,b,c,d>0$
$$\frac{a}{b+2c+d}+\frac{b}{c+2d+a}+\frac{c}{d+2a+b}+\frac{d}{a+2b+c}\geq 1$$

Solution :

$LHS = \displaystyle{\frac{a^2}{ab+2ac+ad}+\frac{b^2}{bc+2bd+ab}+\frac{c^2}{cd+2ac+bc}+\frac{d^2}{ad+2bd+cd}}$

$\geq \displaystyle{\frac{(a+b+c+d)^2}{2(ab+bc+cd+da)+4(ac+bd)}}$

$\geq 1$

4. Romania 2005 $a,b,c>0$ ,$a+b+c\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
$$a+b+c\geq \frac{3}{abc}$$

Solution :

By Cauchy-Schwarz inequality, we have
$$3abc(a+b+c)\leq (ab+bc+ca)^2$$
By assumption we have
$$3(ab+bc+ca)\leq 3abc(a+b+c)\leq (ab+bc+ca)^2.$$
Thus $3\leq ab+bc+ca$.
Therefore, $$\frac{3}{abc}\leq 3\Big(\frac{a+b+c}{ab+bc+ca}\Big)\leq a+b+c.$$