解
1+2x^4-(2x³+x²)
=2x^4-2x³-x²+1
=2x³(x-1)-(x-1)(x+1)
=(x-1)[2x³-x-1]
=(x-1)[x³-x+x³-1]
=(x-1)²[x(x+1)+(x²+x+1)]
=(x-1)²[2x²+2x+1]
=(x-1)²[x²+(x+1)²]≥0
所以
1+2x^4≥2x³+x²
解
1+2x^4-(2x³+x²)
=2x^4-2x³-x²+1
=2x³(x-1)-(x-1)(x+1)
=(x-1)[2x³-x-1]
=(x-1)[x³-x+x³-1]
=(x-1)²[x(x+1)+(x²+x+1)]
=(x-1)²[2x²+2x+1]
=(x-1)²[x²+(x+1)²]≥0
所以
1+2x^4≥2x³+x²