diff --git a/univalent_inverse_power_v2.txt b/univalent_inverse_power_v2.txt
index 56b3ef4..22034a4 100644
--- a/univalent_inverse_power_v2.txt
+++ b/univalent_inverse_power_v2.txt
@@ -665,7 +665,7 @@ a contradiction.  Recall rho = 2^beta, so under (2.1) we have
 
     rho >= 2^{1/4} > 1.                                            (2.2)
 
-The strict inequality rho > 1 in (2.2) is used repeatedly: it is what
+The strict ineuality rho > 1 in (2.2) is used repeatedly: it is what
 lets us send rho^{-n} to zero.
 
 The plan of this part.  The operator L has norm growth rate rho.  A
@@ -727,7 +727,7 @@ pointwise, (2.4) gives
 
     L g_tau = tau (g_tau - 1) <= tau (1 - 1/M_tau) g_tau.
 
-That is a pointwise inequality of continuous functions.  Now use
+That is a pointwise ineuality of continuous functions.  Now use
 positivity and induction.  For any c >= 0, if 0 <= Phi <= c g_tau then
 0 <= L Phi <= c L g_tau <= c tau(1-1/M_tau) g_tau.  Since 1 <= 1*g_tau,
 induction gives
@@ -981,7 +981,7 @@ u |-> 2u, so its integral over [y,2y] is the same for every y > 0.
 Comparing with (2.12) (the case y = 1) gives (2.13).  QED
 
 NORMALIZE.  Divide nu by its positive mass nu(1) to obtain a Borel
-probability P on K.  Inequality (2.13) survives the division because
+probability P on K.  Ineuality (2.13) survives the division because
 both sides are linear in the functional.  Unwinding the definition of
 A_z and of the functional notation, (2.13) says:
 
@@ -1291,7 +1291,7 @@ LEMMA 3.4 (decay far out).  Suppose I(F) < infinity.  Then
 
 Proof.
 
-(a)  Fix z = x+iy.  The submean inequality for the subharmonic (indeed
+(a)  Fix z = x+iy.  The submean ineuality for the subharmonic (indeed
 harmonic) function |1/F|^3 = |F|^{-3} on the disk B(z, y/2), which lies
 in H because every point of it has height at least y/2:
 
@@ -1365,7 +1365,7 @@ So
     |H(s)| <= Y_psi INT_H |F'(z)|^2 |(Delta psi)(F(z))| dA(z)
             <= Y_psi INT_C |(Delta psi)(w)| dA(w),
 
-where the last inequality used injectivity to change variables w = F(z).
+where the last ineuality used injectivity to change variables w = F(z).
 The bound does not depend on s.  Since H(s) >= INT_R 1_K(F(x+is)) dx
 by psi >= 1_K, this is (3.9).  QED
 
@@ -1693,7 +1693,7 @@ factor beta + 2 is positive, so the right side of (4.5) is nonnegative,
 and it vanishes exactly when 4beta = 1.  QED
 
 (4.6) is the engine of the proof.  Hold on to it: Part 5 proves the
-opposite inequality, E J_F(i) <= 0, which will force beta = 1/4, and
+opposite ineuality, E J_F(i) <= 0, which will force beta = 1/4, and
 then Part 6 shows that even beta = 1/4 is impossible.
 
 Now the identification of critical points.
@@ -2341,8 +2341,8 @@ PART 5.  STRICT WEIGHTED TRANSPORT
 ================================================================
 
 Part 4 produced a pairing of saddles with taller peaks.  This part
-turns that combinatorial statement into an inequality between
-integrals, and then into the inequality E J_F(i) <= 0.
+turns that combinatorial statement into an ineuality between
+integrals, and then into the ineuality E J_F(i) <= 0.
 
 The idea in one sentence: change variables through G_F, so that a sum
 over the critical points of a fiber becomes an integral over the target
@@ -2639,7 +2639,7 @@ QED
 
 REMARK (where strictness was actually used).  Note carefully that no
 step used a NUMERICAL gap.  The only place strictness entered was
-(5.13), the strict inequality c_F > 1 on D, which came from part (b) of
+(5.13), the strict ineuality c_F > 1 on D, which came from part (b) of
 LEMMA 4.7, which came from the no-tie condition (g3).  Then STEP 7 used
 the elementary principle "a nonnegative function with zero integral
 vanishes".  A non-strict pairing would have given c_F >= 1, hence