Proof that Recall ≥ F1-score I need to prove mathematically that the value of Recall is always bigger or equal to the F1-score.
The equation would be: $2 \times Precision \times \frac{Recall}{(Precision + Recall)} \le Recall$
Does someone have an idea how I can prove this mathematically?

$$
\begin{array}{rl}\begin{array}{c}\\\\
\rlap{\vphantom{\frac{\frac{1}{1}}{\frac{1}{1}}}\require{HTML}\style{display: inline-block; transform: rotate(270deg)}{\text{truth}}}\end{array}&&
\begin{array}{c|cc}
& \rlap{\text{predictions}}
 \\ & P & N\\ \hline
P & TP & FN \\
N  & FP & TP\\
\end{array}\end{array}$$
You have
$$\begin{array}{rcl}
\text{precision} &=& \frac{TP}{TP+FP} \\
\text{recall} &=& \frac{TP}{TP+FN}
\end{array}$$
Recall and precision are related with both having $TP$ in the numerator.
In terms of $TP$, $FP$ and $FN$ you get for the $F$ score.
$$\begin{array}{}F& =& 2\frac{\frac{TP}{TP+FP} \cdot \frac{TP}{TP+FN}}{ \frac{TP}{TP+FP} +  \frac{TP}{TP+FN} } \\ 
& =& \frac{ TP}{TP+FN} \cdot \underbrace{\frac{2(TP+FN)}{2(TP+FN)+FP-FN}}_{\text{$< 1$ if $FP>FN$} }
\end{array}$$
So it is only true if $FP>FN$ or when recall is larger than precision. This makes sense since the F score is the averag of recall and precision. If precision is smaller than recall then the average will be smaller than recall.
