Alternative proof of a transpose property I am asked to prove; $$(AB)^T=B^TA^T$$ although it is very simple to prove it by the straight forward way, in the exercise I am asked to prove it without using subscripts and sums, directly from the following property of inner product of real vectors:$$\langle A\textbf{x},\textbf{y}\rangle =\langle \textbf{x},A^T\textbf{y}\rangle$$
Where $A$ is an $m\times n$ real matrix, $x \in \mathbb{R}^n$,$y \in \mathbb{R}^m$ 
I don't know how to approach this problem, any suggestions? 

Use nondegeneracy of inner product. I.e if for every vector $x\in\mathbb{R}^n$
$$<x,y_1>=<x,y_2>$$
holds, then $y_1=y_2.$ Fix $y\in\mathbb{R^m}$ and from what you have quoted you have that for every $x$
$$<ABx,y>=<Bx,A^Ty>=<x,B^TA^Ty>$$
and
$$<ABx,y>=<x,(AB)^Ty>.$$
Hence $B^TA^Ty=(AB)^Ty.$ $y$ was arbitrary, hence $B^TA^T=(AB)^T.$
