Let $A,B,C$ be the angles of a triangle. Prove that $$\tan^2(\frac{A}{2})+\tan^2(\frac{B}{2})+\tan^2(\frac{C}{2})<2$$ if and only if $$\tan(\frac{A}{2})+\tan(\frac{B}{2})+\tan(\frac{C}{2})<2$$
Since $A+B+C=\pi$
$$\tan(\frac{A}{2})=\cot (\frac{B+C}{2})=\frac{1-\tan(\frac{B}{2})\tan(\frac{C}{2})}{\tan(\frac{B}{2})+\tan(\frac{C}{2})}$$
then $$\tan(\frac{A}{2})\tan(\frac{B}{2})+\tan(\frac{B}{2})\tan(\frac{C}{2})+\tan(\frac{C}{2})\tan(\frac{A}{2})=1$$
Hence, $\tan^2(\frac{A}{2})+\tan^2(\frac{B}{2})+\tan^2(\frac{C}{2})<2$ is equivalent to $$(\tan(\frac{A}{2})+\tan(\frac{B}{2})+\tan(\frac{C}{2}))^2<4$$
which is also $$\tan(\frac{A}{2})+\tan(\frac{B}{2})+\tan(\frac{C}{2})<2$$
I think your proof is much more better, but we can use also the law of cosines.
We have $$\sum_{cyc}\tan\frac{\alpha}{2}<2$$ it's $$\sum_{cyc}\sqrt{\frac{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}<2$$ or $$\sum_{cyc}\sqrt{\frac{a^2-(b-c)^2}{(b+c)^2-a^2}}<2$$ or $$\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{(a+b+c)(b+c-a)}}<2$$ or $$\sum_{cyc}(a+b-c)(a+c-b)<2\sqrt{(a+b+c)\prod_{cyc}(a+b-c)}$$ or $$\left(\sum_{cyc}(a+b-c)(a+c-b)\right)^2<4(a+b+c)\prod_{cyc}(a+b-c)$$ or $$\sum_{cyc}(a+b-c)^2(a+c-b)^2+2\sum_{cyc}(a+b-c)\prod_{cyc}(a+b-c)<4(a+b+c)\prod_{cyc}(a+b-c)$$ or $$\sum_{cyc}(a+b-c)^2(a+c-b)^2+2\sum_{cyc}a\prod_{cyc}(a+b-c)<4(a+b+c)\prod_{cyc}(a+b-c)$$ or $$\sum_{cyc}(a+b-c)^2(a+c-b)^2<2(a+b+c)\prod_{cyc}(a+b-c)$$ or $$\sum_{cyc}\frac{(a+b-c)(a+c-b)}{(a+b+c)(b+c-a)}<2$$ or $$\sum_{cyc}\frac{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}<2$$ or $$\sum_{cyc}\tan^2\frac{\alpha}{2}<2$$ and we are done!