a+b+c
=(x^2-2x+1)+(y^2+2y+1)+(z^2+2z+1)+(π/2+π/3+π/6)-3
=(x-1)^2+(y+1)^2+(z+1)^2+π-3
因为π-3>0,又平方数大于等于零
所以有:a+b+c>0
即a b c中至少有一个大于0.
若a b c均为实数,且a=x2+2y+π/2 ,b=y2+2z+π/3 ,c=z2-2x+π/6,求证a b c中至少有一个大于0.
若a b c均为实数,且a=x2+2y+π/2 ,b=y2+2z+π/3 ,c=z2-2x+π/6,求证a b c中至少有一个大于0.
数学人气:580 ℃时间:2019-08-22 21:05:45
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