Show that $|uv^T-wz^T|_F^2le |u-w|_2^2+|v-z|_2^2$

Let $A=(u-w)v^T$ and $B=w(v-z)^T$. The inequality in question is then equivalent to $$ |A+B|_F^2le|A|_F^2+|B|_F^2. $$ It is true if and only if $langle A,Brangle_Fle0$. Indeed, this is the case because $$ langle A,Brangle_F =left[w^T(u-w)right]left[v^T(v-z)right] =(w^Tu-1)(1-v^Tz)le0. $$

Another approach: note that for orthogonal matrices $U,V,$ we have $$ |uv^T - wz^T|_F^2 = |U(uv^T - wz^T)V|_F^2 = |(Uu)(Vv)^T - (Uw)(Vz)^T|_F^2. $$ So without loss of generality, we can assume that $u = v = (1,0,dots,0)^T$, so $uv^T$ is the matrix with a $1$ as its $1,1$ entry and zeros elsewhere. The left-hand side is then given by $$ |uv^T - wz^T|_F^2 = |wz^T|_F^2 + [(1 - w_1z_1)^2 - (w_1z_1)^2] \ = (w^Tz)(z^Tw) + [1 - 2w_1 z_1] = 2 - 2w_1z_1. $$ The right hand size is given by $$ |u - w|^2 + |v-z|^2 = |w|^2 + [(1 - w_1)^2 - w_1^2] + |z|^2 + [(1 - z_1)^2 - z_1^2] \ = 2 - 2w_1 + 2 - 2z_1 = 4 - 2(w_1 + z_1), $$ and from there the reasoning is similar.

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Another approach for expanding the exact expression: note that $$ M = uv^T - wz^T = pmatrix{u & w} pmatrix{v & -z}^T, $$ so that $$ operatorname{tr}(MM^T) = operatorname{tr}[pmatrix{u & w} pmatrix{v & -z}^Tpmatrix{v & -z}pmatrix{u & w}^T] \ = operatorname{tr}[pmatrix{v & -z}^Tpmatrix{v & -z}pmatrix{u & w}^Tpmatrix{u & w}] \ = operatorname{tr}left[pmatrix{1 & -v^Tz\ -v^Tz & 1}pmatrix{1 & u^Tw\u^Tw & 1}right] $$

Let $a=u^Tw,b=v^Tz.$

$|uv^T-wz^T|_F^2=tr((uv^T-wz^T)^T(uv^T-wz^T))=2-2ab$

And RHS=$4-2(a+b).$

Check that $2-2able4-2(a+b) iff a+b-able1,$ using that $|a|le1, |b|le 1.$