Philip Ellison	Posted on Wednesday, 03 September, 2003 - 01:30 pm: Prove that, for $a$, $b$, $c\in R^{+}$, $a^2+b^2+c^2\ge \mathrm{ab}+\mathrm{bc}+\mathrm{ca}$. There must be a nice factorisation here, but I can't spot it! Can someone please provide me with a hint.
David Loeffler	Posted on Wednesday, 03 September, 2003 - 01:38 pm: As with any inequality there are all sorts of ways of doing this. The canonical one is to expand out (a-b)2 and get an inequality, and add three copies of it with different sets of variables. If you're feeling like a slightly offbeat approach, apply either the Cauchy-Schwarz inequality or Chebyshev's inequality to (a,b,c) and (c,b,a). The result is true for arbitrary real a,b,c, not necessarily positive, as any of these proofs will show. If you only want a proof for positive values there are even more silly ways. Try and deduce something from 3-variable AM-GM and the factorisation a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca), but remember that the orthodox proofs of AM-GM assume positivity of all the variables. David
Graeme McRae	Posted on Wednesday, 03 September, 2003 - 03:20 pm: Expanding on David's canonical way to prove this, a2 - 2ab + b2 > = 0 (because it's a square), so a2 + b2 > = 2ab. Similarly, b2 + c2 > = 2bc, and c2 + a2 > = 2ca, so a2 + b2 + c2 > = ab + bc + ca
Philip Ellison	Posted on Wednesday, 03 September, 2003 - 03:28 pm: Thanks very much. David: the cubic factorisation you give is the one that I couldn't remember!