已知a属于R,求证:3(1+a^2+a^4)>=(1+a+a^2)^2

已知a属于R,求证:3(1+a^2+a^4)>=(1+a+a^2)^2
已知a属于R,求证:3(1+a^2+a^4)>=(1+a+a^2)^2

已知a属于R,求证:3(1+a^2+a^4)>=(1+a+a^2)^2
3(1+a^2+a^4)-(1+a+a^2)^2
=3+3a^2+3a^4-1-a^2-a^4-2a-2a^2-2a^3
=2+2a^4-2a-2a^3
=2[a^3(a-1)-(a-1)]
=2(a-1)(a^3-1)
=2(a-1)(a-1)(a^2+a+1)
=2(a-1)^2[(a+1/2)^2+3/4]
因为(a+1/2)^2+3/4>0
(a-1)^2>=0
所以2(a-1)^2[(a+1/2)^2+3/4]>=0
所以3(1+a^2+a^4)-(1+a+a^2)^2>=0
所以3(1+a^2+a^4)>=(1+a+a^2)^2

不等式右侧展开式：
a^4 + 2a^3 + 3a^2 + 2a + 1
左-右
=2a^4 - 2a^3 - 2a +2
=2(a^3 - 1)(a - 1)
=2(a - 1)(a^2 + a + 1)(a - 1)
=2(a^2 + a + 1)(a - 1)^2
(a^2 + a + 1)恒正
(a - 1)^2 ≥0
证毕

3(1+a^2+a^4)-(1+a+a^2)^2
=3(1+a^2+a^4)-(1+a^2+a^4+2a+2a^2+2a^3)
=2+2a^2+2a^4-2a-2a^2-2a^3
=(1-2a+a^2)+(1-2a^2+a^4)+(a-2a^3+a^4)
=(1-a)^2+(1-a^2)^2+(a-a^2)^2
≥0
所以，
3(1+a^2+a^4)>=(1+a+a^2)^2

不等式右侧展开式：
a^4 + 2a^3 + 3a^2 + 2a + 1
左-右
=2a^4 - 2a^3 - 2a +2
=2(a^3 - 1)(a - 1)
=2(a - 1)(a^2 + a + 1)(a - 1)
=2(a^2 + a + 1)(a - 1)^2
(a^2 + a + 1)=a^2 + a + 1/4+3/4=(a+1/2)^2+3/4恒正
(a - 1)^2 ≥0
证毕