cho x,y,z >0 thỏa mãn xyz=x+y+z+2.chứng minh rằng:
a,$xy+yz+zx\geq 2(x+y+z)$
b,$\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{3}{2}\sqrt{xyz}$
chứng minh rằng: $xy+yz+zx\geq 2(x+y+z)$
Started By lovemoon, 06-02-2013 - 08:36
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Posted 06-02-2013 - 08:36
lovemoon
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