China Mathematical Competition (Hainan)

Question

Suppose that and are different real roots of the equation 4x^2 - 4tx - 1 = 0 (t R). [, ] is the domain of the function f(x) = 2x-tx^2+1. (1) Find g(t) = f(x) - f(x). (2) Prove that for u_i (0, 2) (i = 1, 2, 3), if u_1 + u_2 + u_3 = 1, then: 1g( u_1) + 1g( u_2) + 1g( u_3) < 346.

Accepted Answer

(1) Let x_1 < x_2 , then 4x_1^2 - 4tx_1 - 1 0, 4x_2^2 - 4tx_2 - 1 0. Therefore, 4(x_1^2 + x_2^2) - 4(t(x_1 + x_2) - 2) 0, 2x_1x_2 - t(x_1 + x_2) - 12 < 0. But f(x_2) - f(x_1) = 2x_2 - tx_2^2 + 1 - 2x_1 - tx_1^2 + 1 = (x_2 - x_1)[t(x_1 + x_2) - 2x_1x_2 + 2](x_2^2 + 1)(x_1^2 + 1), and t(x_1 + x_2) - 2x_1x_2 + 2 > t(x_1 + x_2) - 2x_1x_2 + 12 > 0, thus f(x_2) - f(x_1) > 0. Consequently, f(x) is an increasing function on the interval [, ]. Since + = t and = -14, g(t) = f(x) - f(x) = f() - f() = t^2 + 1(t^2 + 52)t^2 + 2516 = 8t^2 + 1(2t^2 + 5)16t^2 + 25. (2) g( u_i) = 8 u_i(2^2 u_i + 3)16^2 u_i + 9 = 16 u_i + 24 u_i16 + 9^2 u_i 216 2416 + 9^2 u_i = 16616 + 9^2 u_i (i = 1, 2, 3), so align* _i=1^3 1g( u_i) & 1166 _i=1^3 (16 + 9^2 u_i) &= 1166 (16 3 + 9 3 - 9 _i=1^3 ^2 u_i). align* Since _i=1^3 u_…