Nice proof, **Opalg**.

Here is another proof (which I took all night thinking about).

$\displaystyle a\ <\ b+c$

$\displaystyle \implies\ a\ <\ a+b+c-a\ =\ 2-a$

$\displaystyle \implies\ 1-a\ >\ 0$

Similarly $\displaystyle 1-b>0$ and $\displaystyle 1-c>0.$

$\displaystyle \therefore\ (1-a)(1-b)(1-c)\ >\ 0$

$\displaystyle \implies\ 1-(a+b+c)+ab+bc+ca-abc\ >\ 0$

$\displaystyle \implies\ ab+bc+ca-abc\ >\ 1$

Now multiply through by 2, replace $\displaystyle 2(ab+bc+ca)$ by $\displaystyle (a+b+c)^2-(a^2+b^2+c^2)$ and rearrange to get the desired inequality.