Second Squeeze Claim:For any tractions {$F\neq G$}, {$$\mathcal S(FG)\le\frac23\mathcal S(G).$$} Proof: By the First Squeeze Claim, {$\mathcal S(CG)\le \frac12\mathcal S(G)$}. The tractions {$L,R,M$} are similar to each other; consider {$F=R$}. Let {$d=\mathcal S(G)$}. Squeeze Subclaim: If {$\langle a,b,c\rangle$} is in the range of {$G$}, then {$a+b\ge.5$} and {$c\le.5$}. We have: {$$\left|\frac 1{1+a+b} - \frac 1{1+a'+b'}\right|$$} {$$= \left|\frac 1{2-c} - \frac 1{2-c'}\right|$$} {$$=\left|\frac{c'-c}{(2-c)(2-c')}\right|\le \left|\frac{d}{(2-c)(2-c')}\right|$$} Since {$c\le.5$}, {$c'\le.5$}, {$2-c\ge 1.5$}, {$2-c'\ge 1.5$} so we have: {$$\left|\frac 1{1+a+b} - \frac 1{1+a'+b'}\right|\le \frac 49d$$} We also have: {$$\left|\frac a{1+a+b} - \frac {a'}{1+a'+b'}\right|$$} {$$= \left|\frac {a-a'+ab'-a'b}{(1+a+b)(1+a'+b')}\right|$$} {$$=\left|\frac {a-a'+a(b'-b)+b(a-a')}{(1+a+b)(1+a'+b')}\right|$$} {$$\le\left|\frac {d(1+a+b)}{(1+a+b)(1+a'+b')}\right|$$} {$$\le \frac{d}{1+a'+b'}$$} Again, since {$a'+b'\ge.5$} we have: {$$\left|\frac a{1+a+b} - \frac {a'}{1+a'+b'}\right|\le\frac23d.$$} |