Consider the following theorem and proof.
Theorem
Let Γ be a convex and compact set in $R^n$, let f be a strictly positive convex numerical function, and let $F(x)=1/f(x)$ . Then F(x) is convex.
Proof
If we assume that F(x) is convex, by definition of convex function we have
$F\left( {\theta x + \left( {1 - \theta } \right)y} \right) \leqslant \theta F\left( x \right) + \left( {1 - \theta } \right)F\left( y \right)$
For $x,y\, \in \,dom\,F$ and $0 \leqslant \theta \leqslant 1$ By replacing $F(x)=1/f(x)$
$\frac{1}{{f\left( {\theta x + \left( {1 - \theta } \right)y} \right)}} \leqslant \theta \frac{1}{{f\left( x \right)}} + \left( {1 - \theta } \right)\frac{1}{{f\left( y \right)}}$
After some manipulation we have
$f\left( x \right)f\left( y \right) \leqslant f\left( {\theta x + \left( {1 - \theta } \right)y} \right)\left( {\theta f\left( y \right) + \left( {1 - \theta } \right)f\left( x \right)} \right)$
Since $f\left( x \right)$ is a convex function
$\begin{gathered} f\left( x \right)f\left( y \right) \leqslant f\left( {\theta x + \left( {1 - \theta } \right)y} \right)\left( {\theta f\left( y \right) + \left( {1 - \theta } \right)f\left( x \right)} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \leqslant \left( {\theta f\left( x \right) + \left( {1 - \theta } \right)f\left( y \right)} \right)\left( {\theta f\left( y \right) + \left( {1 - \theta } \right)f\left( x \right)} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ = \left( {{\theta ^2} + {{\left( {1 - \theta } \right)}^2}} \right)f\left( x \right)f\left( y \right) + \left( {1 - \theta } \right)\theta \left( {{f^2}\left( y \right) + {f^2}\left( x \right)} \right) \hfill \\ \end{gathered} $
After some manipulation and simplification
$\begin{gathered} {f^2}\left( x \right) + {f^2}\left( x \right) - 2f\left( x \right)f\left( y \right) \geqslant 0 \hfill \\ {\left( {f\left( x \right) - f\left( y \right)} \right)^2} \geqslant 0 \hfill \\ \end{gathered} $
Since this condition is true, hence F(x) is convex.
End of proof
But the above proof does not seem to apply in all cases.
For example: $f\left( x \right) = {x^2} + 1$ .
Any comments or insights as to why this is?
Thanks in advance
The theorem is false, as the case $f(x)=x^2+1$ shows.
You need to read your "proof" backwards, since you wanna show that something true implies $$ F(\theta x+(1-\theta)y) \le \theta F(x) + (1-\theta)F(y). $$ You correctly show that $$ \big[\theta f(x)+(1-\theta)f(y)\big]\big[\theta f(y) + (1-\theta)f(x)\big] \ge f(x)f(y), $$ and that $$ \big[\theta f(x)+(1-\theta)f(y)\big]\big[\theta f(y) + (1-\theta)f(x)\big] \ge f\big(\theta x+(1-\theta)y\big) \big[\theta f(y) + (1-\theta)f(x)\big] , $$ but you can't conlude from this that $$ f\big(\theta x+(1-\theta)y\big) \big[\theta f(y) + (1-\theta)f(x)\big] \ge f(x)f(y), $$ which is what you need.
See also this answer.