[nooonuii]

ุึ67. Proposed for USAMO 1999 $x,y,z>1$
$$x^{x^2+2yz}y^{y^2+2zx}z^{z^2+2xy}\geq (xyz)^{xy+yz+zx}$$

Solution :

WLOG assume $x\geq y\geq z$. Then

$x^{x^2+yz-zx-xy}y^{y^2+zx-xy-yz}z^{z^2+xy-yz-zx}$

$\geq y^{x^2+yz-zx-xy}y^{y^2+zx-xy-yz}z^{z^2+xy-yz-zx}$

$=y^{x^2+y^2-2xy}z^{z^2+xy-yz-zx}$

$\geq z^{x^2+y^2-2xy}z^{z^2+xy-yz-zx}$

$=z^{x^2+y^2+z^2-xy-yz-zx}$

$\geq 1.$

68. Turkey 1999 $c\geq b\geq a \geq 0$
$$(a+3b)(b+4c)(c+2a)\geq 60abc$$

Solution :

We prove the following inequalities

$(1) \, 4ac^2 + 2bc^2 + 6ab^2 \geq 12abc$

$(2) \, 8bc^2 + 8a^2c\geq 16abc$

$(3) \, 2bc^2 + 2a^2b \geq 4abc$

$(4) \, 3b^2c \geq 3abc$

$(5) \, 25abc \geq 25abc.$

Adding them together we get the result.

70. Poland 1999 $a,b,c>0,a+b+c=1$
$$a^2+b^2+c^2+2\sqrt{3abc}\leq 1$$

Solution :

Since $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)=1-2(ab+bc+ca)$, the inequality is equivalent to
$$3abc\leq (ab+bc+ca)^2.$$

But this inequality is just $3abc(a+b+c)\leq (ab+bc+ca)^2$ which can be shown by noting that

$$3\Big(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\Big)\leq (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2. $$