proposition: Lemma sqrt_more : forall x:R, 0 < x -> x < 1 -> x < sqrt x.
proof: Proof. intros x H1 H2; generalize (sqrt_lt_1 x 1 (Rlt_le 0 x H1) (Rlt_le 0 1 Rlt_0_1) H2); intro H3; rewrite sqrt_1 in H3; generalize (Rmult_ne (sqrt x)); intro H4; elim H4; intros H5 H6; rewrite <- H5; pattern x at 1; rewrite <- (sqrt_def x (Rlt_le 0 x H1)); apply (Rmult_lt_compat_l (sqrt x) (sqrt x) 1 (sqrt_lt_R0 x H1) H3). Qed.
imports: Rbase Rfunctions Rsqrt_def