I have the following inequality:
$$ f(x)^T(f(x) - A x) \leq 0 $$
The $A$ is a positive definite matrix of dimension $a \times a$, $x$ is a $a \times 1$ vector and $f$ is a vector valued function that has $a$ dimensions (everything real valued). Why follows from this:
$$ f(x)^T(A^{-1}f(x) - x) \leq 0 $$
I can't multiply $A^{-1}$ from the left side because the $f(x)^T$ is in the way...
This is not true. E.g. let $$ x=\pmatrix{1\\ 1},\ f(x)=\pmatrix{1\\ -1},\ A=\pmatrix{4&0\\ 0&1}. $$ Then \begin{aligned} f(x)^T(f(x)-Ax)&=\pmatrix{1&-1}\left[\pmatrix{1\\ -1}-\pmatrix{4\\ 1}\right] =\pmatrix{1&-1}\pmatrix{-3\\ -2}=-1<0,\\ f(x)^T(A^{-1}f(x)-x)&=\pmatrix{1&-1}\left[\pmatrix{\frac14\\ -1}-\pmatrix{1\\ 1}\right] =\pmatrix{1&-1}\pmatrix{-\frac34\\ -2}=\frac54>0. \end{aligned}