Suppose there is such an a, b, c b, c.

Then: from f( 1)=7/2:

a+b+c=7/2，

a=7/2-b-c

Due to:

f(x)≤2x^2+2x+(3/2)

That is, ax 2+bx+c ≤ 2x 2+2x+(3/2)

It applies to all real numbers x,

There are:

(7/2-b-c-2)x^2+(b-2)x+c-3/2≤0

So:

( 1)7/2-b-c-2=0，b-2=0，c-3/2≤0，

Then: b=2, c=- 1/2, a=2.

X 2+(1/2) ≤ f (x)

x2+( 1/2)≤2x 2+2x- 1/2。

For x, not everything is true.

(2)7/2-b-c-2 & lt； 0,

And delta = (b-2) 2-4 (3/2-b-c) (c-3/2) ≤ 0.

b+ c & gt； 1.5①,

b^2+4bc+4c^2- 10b- 12c+ 13≤0②

x ^ 2+( 1/2)≤f(x)

That is, x2+(1/2) ≤ (7/2-b-c) x2+bx+c.

This holds true for all real numbers x.

There is (5-2b-2c) x 2+2bx+2c- 1 ≥ 0.

Same 5-2b-2c >0,

△=(2b)^2-4(5-2b-2c)(2c- 1)≤0

b+ c & lt； 2.5③,

b^2+4bc+4c^2-2b- 12c+5 ≤0④

Obviously, b=c= 1, and a=3/2 satisfies the above category.