If m, n ∈ [- 1, 1], M + n ≠ 0, then f (m) + F (n) m + n & gt; 0. (1) judge the monotonicity of F (x) on [- 1, 1], and prove your conclusion; (2) solve the inequality: F (x + 12) & lt; F (1 x − 1); (3) if f (x) ≤ t 2-2 at + 1 holds for all x ∈ [- 1,1], a ∈ [- 1,1], find the value range of real number t

If m, n ∈ [- 1, 1], M + n ≠ 0, then f (m) + F (n) m + n & gt; 0. (1) judge the monotonicity of F (x) on [- 1, 1], and prove your conclusion; (2) solve the inequality: F (x + 12) & lt; F (1 x − 1); (3) if f (x) ≤ t 2-2 at + 1 holds for all x ∈ [- 1,1], a ∈ [- 1,1], find the value range of real number t

(1) Let - 1 ≤ X1 & lt; x2 ≤ 1, then f (x1) - f (x2) = f (x1) + F (- x2) = f (x1) + F (- x2) x1-x2 · (x1-x2) ∵ - 1 ≤ X1 & lt; x2 ≤ 1, ∵ X1 + (- x2) ≠ 0, from the known f (x1) + F (- x2) x1-x2 & gt; 0, then x1-x2 & lt; 0, ∵ f (x1) - f (x2) & lt; 0, that is, f (x) is an increasing function on [- 1,1]; (2) ∵ f (x) is an increasing function on [- 1,1], so there is - 1 ≤ x + 12 ≤ 1-1 ≤ lx-1 ≤ LX + 12 & lt; 1x-1, from which {x | - 32 ≤ X & lt; -1} (3) it can be seen from (1): F (x) is an increasing function on [- 1,1], and f (1) = 1, so for X ∈ [- 1,1], f (x) ≤ 1. Therefore, to make f (x) ≤ t2-2at + 1, for all x ∈ [- 1,1], a ∈ [- 1,1] hold constant, that is to say, t2-2at + 1 ≥ 1 holds, so t2-2at ≥ 0 holds. That is, G (a) = t2-2at holds constant for a ∈ [- 1,1], G (a) ≥ 0, only G (a) on [- 1,1] So t & gt; 0g (1) ≥ 0 or t ≤ 0g (- 1) ≥ 0. The solution is t ≤ - 2 or T = 0 or t ≥ 2