Proof: Condition for a function being strongly convex Let $\lambda \in \mathbb{R}_{> 0}$. A function $f : \mathbb{R}^d \to \mathbb{R}$ is $\lambda$-strongly convex, if for all $\alpha \in (0, 1)$ and all $u, v \in \mathbb{R}^d$
$$
f(\alpha u + (1 - \alpha) v) \leq \alpha f(u)  + (1 - \alpha) f(v) - \frac{\lambda}{2} \alpha (1 - \alpha) \lVert u - v\rVert^2
$$ holds.
In this blog post (ii -> iv) I found a condition for $f$ being $\lambda$-strongly convex:

If $g(x) = f(x) - \frac{\lambda}{2} \lVert x \rVert^2$ is convex, then $f$ is $\lambda$-strongly convex.

The author claims that this condition follows directly from definition of $g$ and convexity, however I'm having problems of proofing it for myself. This is basically how far I got:
$$\begin{align*}
 &\alpha f(u) + (1 - \alpha) f(v) - \frac{\lambda}{2} \alpha (1 - \alpha) \lVert{u - v}\rVert^2 \\
        &= \alpha f(u) + (1 - \alpha) f(v) - \frac{\lambda}{2} \alpha (1 - \alpha) \left({\lVert{u}\rVert^2 + \lVert{v}\rVert^2 - 2\langle u, v\rangle}\right) \\
        &\geq \alpha f(u) + (1 - \alpha) f(v) - \frac{\lambda}{2} \alpha (1 - \alpha) \left({\lVert{u}\rVert^2 + \lVert{v}\rVert^2}\right) \\
        &= \alpha f(u) + (1 - \alpha) f(v) - \frac{\lambda}{2} \alpha (1 - \alpha) \lVert{u}\rVert^2 - \frac{\lambda}{2} \alpha (1 - \alpha) \lVert{v}\rVert^2 \\
        &> \alpha f(u) + (1 - \alpha) f(v) - \frac{\lambda}{2} \alpha \lVert{u}\rVert^2 - \frac{\lambda}{2} (1 - \alpha) \lVert{v}\rVert^2 \\
        &= \alpha \left({ f(u) - \frac{\lambda}{2} \lVert{u}\rVert^2 }\right) + (1 - \alpha) \left({ f(v) - \frac{\lambda}{2} \lVert{v}\rVert^2 }\right) \\
        &= \alpha g(u) + (1 - \alpha) g(v) \\
        &\geq g(\alpha u + (1 - \alpha) v) \\
        &= f(\alpha u + (1 - \alpha) v) - \frac{\lambda}{2} \lVert{\alpha u + (1 - \alpha) v}\rVert^2 \\
        &= ? \\
        &= f(\alpha u + (1 - \alpha) v)\end{align*}$$
What am I missing?
 A: Thanks to @whuber I noticed that especially the $-2 \langle u, v \rangle$-term is to valueable to discard. I restarted the calculation from the other side and was able to show the required bound by using $\leq$ exactly one time (applying the premise).
In case anybody else is interested in a sketch of the solution:
Start by showing
$$
\begin{align*}
&f(\alpha u + (1 - \alpha) v) \\
&= f(\alpha u + (1 - \alpha) v) - \frac{\lambda}{2} \lVert{\alpha u + (1 - \alpha) v}\rVert^2 + \frac{\lambda}{2} \lVert{\alpha u + (1 - \alpha) v}\rVert^2 \\
&= \dots \\
&\leq \alpha f(u) + (1 - \alpha) f(v) - \frac{\lambda}{2} \alpha \lVert{u}\rVert^2 - \frac{\lambda}{2} (1 - \alpha) \lVert{v}\rVert^2 + \frac{\lambda}{2} \lVert{\alpha u + (1 - \alpha) v}\rVert^2,
\end{align*}
$$
and then conclude using only basic linear algebra that 
$$
- \frac{\lambda}{2} \alpha \lVert{u}\rVert^2 - \frac{\lambda}{2} (1 - \alpha) \lVert{v}\rVert^2 + \frac{\lambda}{2} \lVert{\alpha u + (1 - \alpha) v}\rVert^2
= -\frac{\lambda}{2} \alpha (1 - \alpha) \lVert{u - v}\rVert^2.
$$
