Sauron

$GT$ => $xy$ $\leq$ $4$

$P$ = $\frac{2010}{xy}$+$\frac{12}{x^{2}+y^{2}}$+$9xy$ =($\frac{12}{x^{2}+y^{2}}$+$\frac{12}{2xy}$)+($9xy+\frac{144}{xy}$)+$\frac{1860}{xy}$ $\geq$ $12(\frac{4}{(x+y)^{2}})+2\sqrt{9xy.\frac{144}{xy}}+\frac{1860}{4}$ $\geq$ $540$

$"="$ xảy ra : $x=y=2$

$\sum\frac{x^2}{x^2+y^2-xy} = \sum\frac{x^2(x+y)}{x^3+y^3} = \sum\frac{x^3+x^2y}{x^3+y^3}= 3+\sum\frac{x^2y-y^3}{x^3+y^3} \leq 3+\sum\frac{y(x-y)(x+y)}{xy(x+y)}=3+\sum\frac{x-y}{x}=6-\frac{y}{x}-\frac{z}{y}-\frac{x}{z}\leq 3$.

Cho số nguyên dương x,y,z thỏa: $x^{2} +y^{2}=z^{2}.$

CM $xyz$ $\vdots$ $3$ và $\vdots$ $60$