You said you have proved it in the one variable case but in my opinion it should be wrong already in that case: Take $L_1=1+X$, $L_2=1-X$ and $P=X^2$. Then each $L_i$ is linear, $P$ is quadratic and $P\ge0$ on $\{L_i\ge0\}$. On the other hand each product of some of the $L_i$ is positive at the origin, showing that the desired representation cannot exist.

If you allow however for $P$ being strictly positive, then you have much better chances. You might want to have a look at the following books...

 - Marshall: Positive polynomials and sums of squares
 - Prestel: Positive polynomials
 - Bochnak, Coste, Roy: Real algebraic geometry
 - Basu, Pollack, Roy: Algorithms in real algebraic geometry
 - Knebusch, Scheiderer: Einführung in die reelle Algebra
 - Andradas, Bröcker, Ruiz: Constructible sets in real geometry

...and the following articles...

- http://homepages.cwi.nl/~monique/files/moment-ima-update-new.pdf
- http://www.math.uni-konstanz.de/~schweigh/publications/sosdualsdp.pdf

Also the so-called "S-procedure" could be of interest for you.