数学归纳法不等式,求n最小值,3^n>n^4

n=7
3^nn=8
3^n>n^4
设n=k时,k>=8
有3^k>k^4
则n=k+1时
3^(k+1)=3*3^k>3k^4
3k^4-(k+1)^4
=3k^4-(k²+2k+1)²
=(√3k²-k²-2k-1)(√3k²+k²+2k+1)
√3k²-k²-2k-1
=(√3-1)[k-1/(√3-1)]²-1/(√3-1)-1
=(√3-1)[k-(√3+1)/2]²-(√3+1)/2-1
=(√3-1)[k-(√3+1)]²-(√3+1)/2-1
k>=8,则定义域在对称轴k=(√3+1)/2右边, 是增函数
k=8,√3k²-k²-2k-1=64√3-81>0
所以k>=8,√3k²-k²-2k-1>0
√3k²+k²+2k+1
=(√3+1)[k+1/(√3+1)]²-1/(√3+1)+1
=(√3+1)[k-(-√3+1)/2]²-(√3+1)/2-1
=(√3+1)[k-(-√3+1)/2]²-(√3+1)/2-1
k>=8,则定义域在对称轴k=√3+1右边, 是增函数
k=8,√3k²+k²+2k+1=64√3+81>0
所以k>=8,√3k²+k²+2k+1>0
所以3k^4-(k+1)^4=(√3k²-k²-2k-1)(√3k²+k²+2k+1)>0
即n=k+1,3^(k+1)>(k+1)^4
综上
n>=8,3^n>n^4
所以n最小=8