$$\mathrm{d}U<T\,\mathrm{d}S-p\,\mathrm{d}V= \left(\frac{\partial U}{\partial S}\right)_{\!V}\,\mathrm{d}S +\left(\frac{\partial U}{\partial V}\right)_{\!S}\,\mathrm{d}V$$
So for a change to occur at a constant entropy and volume, the internal energy must decrease. I get that this statement isn't useful because you can't control entropy in a lab. Furthermore, given that: $$\mathrm{d}G<V\,\mathrm{d}p-S\,\mathrm{d}T= \left(\frac{\partial G}{\partial p}\right)_{\!T}\,\mathrm{d}p +\left(\frac{\partial G}{\partial T}\right)_{\!p}\,\mathrm{d}T$$ this shows that a spontaneous change at a constant pressure and temperature is accompanied by a decrease in Gibbs free energy – far more useful as these variables can be controlled.
However, since, for a spontaneous process, $\mathrm{d}U$ must decrease at a constant entropy and volume which is expressed by $\mathrm dU<T\,\mathrm dS-p\,\mathrm dV$. However, expressions for $\mathrm dS$ and $\mathrm dV$ can be expressed as a function of $p$ and $T$ making this inequality a function of pressure an temperature. \begin{align} \mathrm{d}U &<T\,\mathrm{d}S-p\,\mathrm{d}V\\ S&=S(p,T)\\ V&=V(p,T)\\ \end{align}
Sub in the total derivatives and factorize: $$\mathrm{d}U<\left[T\left(\frac{\partial S}{\partial T}\right)_{\!p} - p\left(\frac{\partial V}{\partial T}\right)_{\!p}\right]\,\mathrm{d}T + \left[T\left(\frac{\partial S}{\partial p}\right)_{\!T} - p\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\,\mathrm{d}p$$ Thus, $U=U(p,T)$
This implies that for a spontaneous change at constant pressure and temperature the internal energy must decrease however this is not the case clearly. Does making the substitutions invalidate the inequality? Where is the fault in my reasoning?
So, what's the need for Gibbs free energy. Surely a spontaneous change at constant pressure and temperature is accompanied by a decrease in internal energy? There must be a fault in my reasoning, but where and why?