Can I do the same things on both sides when I have a not equal sign? I am trying to prove this statement:

If $A$ is not symmetric then $A^{-1}$ is not symmetric 

And the following is my proof
If $A≠A^T$,
\begin{align*}
& A^{-1}A=I≠A^{-1} A^T \\
& I^T \neq \left(A^{-1} A^T \right)^T \\
& I \neq A \left( A^{-1} \right)^T \\
& A^{-1} \neq \left( A^{-1} \right)^T \\
& A^{-1} \neq \left( A^{-1} \right)^T. \\
\end{align*}
I am doing several operations on both sides which usually is justified when there is a equal sign between RHS and LHS. I am not sure if this is valid when there is NOT equal sign. 
 A: That way is fine because you are using not equality which is an inequality and when you deal with these it is still fundamental to do to both sides of the equation what you do to one side of the equation
A: As @Daniel Fischer said, this happens to be okay the way you're using it, but it doesn't work in general. See the following counterexample:
$$ \begin{gather*} -1 \neq 1 \\
(-1)^2 \neq (1)^2 \\
1 \neq 1 \\
\text{uh-oh}
 \end{gather*} $$

In particular, applying an operation $f$ to both sides of an inequality will always preserve the inequality if and only if $f^{-1}$ exists. A proof follows.
First we'll show that $f^{-1}$ existing guarantees that the inequality will be held. Take $f$ such that $f^{-1}$ exists and also take $a$ and $b$ such that $ a \neq b $. We now want to show that $f(a) \neq f(b)$. For contradiction, let's assume that $f(a) = f(b)$. Since $f^{-1}$ exists, as we assumed, we may apply it to both sides, giving $f^{-1}(f(a)) = f^{-1}(f(b))$, which is just to say that $a = b$. However, this is a contradiction with $a \neq b$. Thus, assuming that $f(a)=f(b)$ produced a contradiction, so we conclude that $f(a) \neq f(b)$.
Note also that $f^{-1}$ not existing guarantees that inequality won't always be preserved. This is because if $f^{-1}$ doesn't exists, then there are two values $a$ and $b$ such that $a \neq b$ but $f(a) = f(b)$. These two values are an immediate example of when the inequality doesn't hold: though they are unequal, $f$ makes them equal.
A: I think you may do it this time considering |A|≠0 given the fact that there exists A^(-1).
However...Why not prove this： "A^(-1) is symmetric -> A is symmetric" ?

An example of when doing this may cause error.
Let A=$\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right)$,B=$\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 2 \\
\end{array}
\right)$,C=$\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 0 \\
\end{array}
\right)$
Then it's easy to see that A≠B while A·C=B·C
