<s>[INST] Upper bound: Given $L$-smooth convex $f$; $( y- x)^T \left( \nabla f(z)-\nabla f(x)\right)\leq(L/2) ( \| x-z\|^2+\| x-y\|^2+\| z-y\|^2)$? Given $L$-smooth convex $f$, I would highly appreciate if you can confirm whether the following bound is correct or not.
\begin{align}
\left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right) 
\leq \frac{L}{2}  \left( \| x-z\|^2+\| x-y\|^2+\| z-y\|^2 \right) \tag{$\clubsuit$}.
\end{align}

Attempt:
Since function $f$ is both $L$-smooth and convex, I particularly make use of the following two inequalities (e.g., can be found here or many other books such as in Yurii Nesterov's book).

*

*Three points descent lemma:
\begin{align}
f(x) \leq f(y) + \left( x - y \right)^T \nabla f(z) + \frac{L}{2} \| x - z \|^2 
\end{align}

*$$0 \leq f(y) - f(x) - \left( y - x\right)^T \nabla f(x) \leq \frac{L}{2} \| x - y \|^2 $$
To this end, we rewrite the above two respective inequalities
1.
\begin{align}
f(x) &\leq f(y) + \left( x - y \right)^T \nabla f(z) + \frac{L}{2} \| x - z \|^2 \\
\Longleftrightarrow  \left( y - x \right)^T \nabla f(z) &\leq f(y) - f(x) + \frac{L}{2} \| x - z \|^2  \tag{1}
\end{align}
2.
\begin{align}
f(y) - f(x) - \left( y - x\right)^T \nabla f(x) &\leq \frac{L}{2} \| x - y \|^2 \\
\Longleftrightarrow \left( y - x\right)^T \nabla f(x) &\geq f(y) - f(x) - \frac{L}{2} \| x - y \|^2 \tag{2} 
\end{align}
Now,
\begin{align}
\left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right) 
=& \underbrace{\left( y- x\right)^T \nabla f(z)}_{ \text{upper bound using} \ (1) } - \underbrace{\left( y- x\right)^T  \nabla f(x)}_{ \text{lower bound using} \ (2) } \\
=& \underbrace{\left( y- x\right)^T \nabla f(z)}_{ \leq f(y) - f(x) + \frac{L}{2} \| x - z \|^2 }   
- \underbrace{\left( y- x\right)^T  \nabla f(x) }_{ \geq f(y) - f(x) - \frac{L}{2} \| x - y \|^2 } \\
\leq & f(y) - f(x) + \frac{L}{2} \| x - z \|^2 - \left[ f(y) - f(x) - \frac{L}{2} \| x - y \|^2 \right] \\
=& \frac{L}{2} \left(  \| x - z \|^2  + \| x - y \|^2  \right) \\
\leq& \frac{L}{2} \left(  \| x - z \|^2  + \| x - y \|^2 + \underbrace{\| z - y \|^2}_{\geq 0} \right)
\end{align}
This completes the proof of $(\clubsuit)$.
 [/INST] What about this simple estimate:
$$
\left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right) 
\le \|y-x\| \cdot \|\nabla f(z)-\nabla f(x)\| 
\le L \|y-x\| \cdot \|z-x\|
\le \frac L2(\|y-x\|^2 + \|z-x\|^2)
$$
??
</s>