When x & gt=a

f(x)=2x^2+(x-a)^2=3x^2-2ax+a^2

f(0)=a^2>； = 1

a & lt=- 1

When x

f(x)=2x^2-(x-a)^2=x^2+2ax-a^2

f(0)=-a^2>； = 1

No answer

So a < =- 1

When x & gt=a

f(x)=3x^2-2ax+a^2

When a & lt=0

When x takes a/3, the minimum f(x) is 2/3a 2, and a = 0.

When a & gt=0

When x takes a, the minimum f(x) is 2a 2, and a = 0.

f(x)=0

When x

f(x)=x^2+2ax-a^2

When a & gt=0

When x takes -a, the minimum value of f(x) is-2a 2.

When a & lt=0

When x takes a, the minimum f(x) is 2a 2, and a = 0.

Therefore, the minimum value of f(x) is-2a 2, a >；; 0

x & gta,h(x)=3x^2-2ax+a^2>； = 1

If a 2 >; =3/2

The solution set is (a, +∞)

If a 2

The solution set is [(-∞, (a-√3-2a 2)/3)∩((a+√3-2a 2)/3, +∞)] ∩ (a,+∞).