16/x^2+4/(4-x^2)的最小值
问题描述：
16/x^2+4/(4-x^2)的最小值
答
1/a+1/b+1/c>=9/(a+b+c)应用公式16/x^2+4/(4-x^2)=8/x^2+8/x^2+4/(4-x^2)=1/(x^2/8)+1/(x^2/8)+1/((4-x^2)/4)>=9/(x^2/8+x^2/8+((4-x^2)/4)=9所以最小值是9当且仅当x^2/8=((4-x^2)/4)也就是x^2=8/3时成立...