For reasons that probably don't bear examination (I've rewritten my answer to this question, but I haven't posted the new improved version with added vitamins, because I wish to supplement my irritatingly long and pointless school-geometry proof with an equally quixotic school-trigonometry proof, avoiding the theory of convexity), I wish to prove: $$ \text{If } 0 < \alpha < \alpha + \delta < \beta < \frac\pi2 \text{ then } \tan\alpha + \tan\beta > \tan(\alpha + \delta) + \tan(\beta - \delta) $$ using only trigonometric identities familiar from secondary school mathematics.
All I have managed to come up with so far is this pathetic dog's dinner of a proof:
If $0 < \theta < \theta + \varphi = \frac\pi2,$ then $$ \tan\theta + \tan\varphi = \tan\theta + \frac1{\tan\theta} = 2 + \left(\frac1{\sqrt{\tan\theta}} - \sqrt{\tan\theta}\right)^2, $$ which decreases strictly with $\theta$ for $\theta \leqslant \frac\pi4.$
If $\alpha + \beta = \frac\pi2,$ then either $\alpha + \delta \leqslant \frac\pi4$ or $\beta - \delta \leqslant \frac\pi4,$ therefore $\tan\alpha + \tan\beta > \tan(\alpha + \delta) + \tan(\beta - \delta).$ $\ \square$
Assume from now on that $\alpha + \beta \ne \frac\pi2.$ Then $\tan\alpha\tan\beta \ne 1,$ and similarly $\tan(\alpha + \delta)\tan(\beta - \delta) \ne 1.$
Because $\alpha + \beta = (\alpha + \delta) + (\beta - \delta),$ the addition formula for the tangent function gives: \begin{equation} \label{eq:1}\tag{$1$} \frac{\tan(\alpha + \delta) + \tan(\beta - \delta)}{\tan\alpha + \tan\beta} = \frac{1 - \tan(\alpha + \delta)\tan(\beta - \delta)}{1 - \tan\alpha\tan\beta}. \end{equation}
Because $\tan\alpha\tan\delta < 1,$ we can write, after some tedious manipulation (yawn-making details provided on request): $$ \tan(\alpha + \delta)\tan(\beta - \delta) - \tan\alpha\tan\beta = \frac{\tan\delta(1 - \tan^2\alpha\tan^2\beta)[\tan(\beta - \alpha) - \tan\delta]} {(1 - \tan\alpha\tan\delta)(1 + \tan\beta\tan\delta)}. $$ Therefore $\tan(\alpha + \delta)\tan(\beta - \delta) - \tan\alpha\tan\beta$ has the same sign as $1 - \tan\alpha\tan\beta.$
It follows that the right hand side of \eqref{eq:1} is always strictly less than $1,$ and we are done (more like done in!). $\ \square$
Surely it must be possible to do better than this? Please put me out of my misery, so that I can be rid of this stupid obesssion, and perhaps even get on with some slightly more respectable mathematics instead!
We need to prove that if $\{a,b,c,d\}\subset\left(0,\frac{\pi}{2}\right)$, $a\geq b$, $c\geq d$, $a\geq c$ and $a+b=c+d$ so $$\tan{a}+\tan{b}\geq\tan{c}+\tan{d}.$$ Indeed, we need to prove that: $$\frac{\sin(a+b)}{\cos{a}\cos{b}}\geq\frac{\sin(c+d)}{\cos{c}\cos{d}}$$ or $$\cos(c+d)+\cos(c-d)\geq\cos(a+b)+\cos(a-b)$$ or $$a-b\geq c-d$$ or $$a-c\geq b-d,$$ which is obvious because $a-c\geq0$, but $b-d\leq0.$
I think it is better to use Karamata.
Indeeed, if $\beta-\delta\geq\alpha+\delta$ so $$(\beta,\alpha)\succ(\beta-\delta,a+\delta).$$ If $\beta-\delta\leq\alpha+\delta$ so $$(\beta,\alpha)\succ(\alpha+\delta,\beta-\delta).$$ In any case your inequality follows from Karamata because $\tan$ is a convex function on $\left(0,\frac{\pi}{2}\right).$