diff --git a/metric_k_median_threshold_v2.txt b/metric_k_median_threshold_v2.txt
index 0e23278..d2f3640 100644
--- a/metric_k_median_threshold_v2.txt
+++ b/metric_k_median_threshold_v2.txt
@@ -143,7 +143,7 @@ construct a metric instance:
         d(u, v)     = 2   for distinct u, v in U,
         d(A_a, u)   = 1   if u belongs to A_a, and 3 otherwise.
 
-These distances satisfy the triangle inequality. Sample checks:
+These distances satisfy the triangle ineuality. Sample checks:
     d(A_a, A_b) = 2 <= d(A_a, u) + d(u, A_b) = 1 + 1 or 3 + 3.
     d(A_a, u)   <= 3 <= d(A_a, A_b) + d(A_b, u) = 2 + 1 or 2 + 3.
     d(u, v)     = 2 <= d(u, A_a) + d(A_a, v) = 1 + 1 or 1 + 3.
@@ -576,7 +576,7 @@ Let G be the event
 
   sum_s (Z_s - Z_s^*)  <=  gamma * sum_s c_s.
 
-By Markov's inequality:
+By Markov's ineuality:
 
   Pr(G^c) <= [sum_s E(Z_s - Z_s^*)] / (gamma sum_s c_s)
            <= [4g (1+gamma) sum_s c_s] / (gamma sum_s c_s)
@@ -711,7 +711,7 @@ mass on F_j intersect all bundles.
     the difference of Mbar_P from sum_t a_t is bounded by
     (gamma + sqrt(g(1+gamma))) / p_0.
 
-By triangle inequality and by y(U intersect F_j) + sum_t a_t = 1:
+By triangle ineuality and by y(U intersect F_j) + sum_t a_t = 1:
 
   E(1 - M_U - M_P)_+ <= H := sqrt(g) + sqrt(g/p_0)
                             + (gamma + sqrt(g(1+gamma)))/p_0.  (SHORT)
@@ -775,7 +775,7 @@ inputs" statement of Theorem 7.1.
 =============================
 
 The cost analysis in Section 10 will invoke one abstract algebraic
-inequality about a finite probability space. The inequality itself
+ineuality about a finite probability space. The ineuality itself
 says nothing about distances, rounding, or metrics. It is the engine
 that makes the constant 1 + 2/e appear, rather than 2.
 
@@ -786,7 +786,7 @@ The three analytic functions. For x in [0, 1], set
   B(x) = 2 P(x),   m(x) = A(x) - 1 - x B(x).
 
 Key facts (from differentiating, integrating by parts, and elementary
-inequalities; each follows from the integral forms):
+ineualities; each follows from the integral forms):
 
   A'(x) = B(x),
   A(1) = 1 + 2/e,   A(0) = 1,
@@ -800,12 +800,12 @@ Change variable and integrate by parts; the x B(x) term cancels
 the x-dependence cleanly to give 3x. The identity is used in the
 proof of Section 10; we will not retrace it here.
 
-The finite comparison inequality. Fix a finite index set I with
+The finite comparison ineuality. Fix a finite index set I with
 positive weights pi_i summing to 1 (so summation is a probability
 space). Write E_i and E_t for expectation over two independent
 samples from pi.
 
-THEOREM 8.1 (finite comparison inequality). Let 0 <= x <= 1,
+THEOREM 8.1 (finite comparison ineuality). Let 0 <= x <= 1,
 0 < y <= 1, and D_i >= 0 with E_i D_i = 1. Put
 Theta = P(x) (y - x)^2 / y. Suppose q_{ti} in [0, 1] and R_i >= 0
 satisfy the COMPARISON HYPOTHESIS:
@@ -1058,7 +1058,7 @@ Good.
 
 Case otherwise (chi = 0 or s_t = 1 or...):
 The incoming charge is at most H_1 = .25 if there's any charge (in
-the "not covered" case). Apply the scalar-type-one inequality
+the "not covered" case). Apply the scalar-type-one ineuality
 (S5):
   C - k_s + 1.6 z - b >= .25.
 Combining with the trivial q <= 1:
@@ -1092,8 +1092,8 @@ So E_i F_i <= 0. Since E_i[(.4 D_i + 1.6) z] = 2z, this proves
 9. SCALAR BOUNDS
 =============================
 
-Section 8 needed five concrete inequalities on numbers between 0
-and 1. Those inequalities are now stated as a lemma, whose proof is
+Section 8 needed five concrete ineualities on numbers between 0
+and 1. Those ineualities are now stated as a lemma, whose proof is
 in Appendix A.
 
 LEMMA 9.1 (scalar bounds). Fix x in [0, 1], put X = 1 - x. Take
@@ -1117,7 +1117,7 @@ For q, r in [0, 1], set:
   M(r) = (1 - r) a + C P(x) r,
   G(r) = M(r) - L(r).
 
-Then the following five inequalities hold on their stated domains:
+Then the following five ineualities hold on their stated domains:
 
   (S1) m(x) + b - .4 z / r <= C(1 - P(x))/(1 - P(x) r)
                                                  for 0 < r <= 1.
@@ -1155,7 +1155,7 @@ Proofs of (F1)-(F4) are by direct computation:
     (1-q)^2, 1-Pq > 0.
   - (F4): psi(q) = (1-q)/(1-Pq) <= 1-q+Pq because psi(q) <= 1.
 
-The five inequalities (S1)-(S5) are the content of Appendix A. They
+The five ineualities (S1)-(S5) are the content of Appendix A. They
 are proven there by exact rational certificates: interpolate P, A
 with rational polynomials with error <= 7.5e-6; reduce each of
 (S1)-(S5) to "some polynomial in X is nonneg on [0, 1]"; verify