... ≤ α +x 1+ x 1+ α x +1 x +1 2x (x + 1) ⇔ x + x + ≥ (2 + 2α + ) α +x 1+ α x +1 2x (x + 1) * Xét hàm s f (x ) = x + x + − (2 + 2α + ), α +x 1+ α 2(2x + 1) α 1 * Ta có: f '(x ) = 2x + − −2 α +1 (x + ... 2, ∀x ∈ 0 ;1 ( ) * Xét hàm f x = n + x + n − x , x ∈ [0 ;1) 1 ⇒f' x = nn 1+ x ( ) ( − n 1 ) n n 1 (1 − x ) < 0, ∀x ∈ 0 ;1 ( ) 31 ( ) ( ) ( ) () ( ) V y f x gi m 0 ;1 nên f x ... c 1 ⇔ + + ≥ 1+ x 1+ y 1+ z * Gi s z ≤ ⇒ xy ≥ nên có: ⇒ 1 2 z + ≥ = + x + y + xy + z 1 z 2t + + ≥ + = + = f (t ) v i + x + y + z + z + z + t + t2 t = z 1 * Ta có: f '(t ) = ⇒ f (t ) ≥ f (1) = (1...