There is probably a more conventional solution, but back when we discussed the problems here's something I found while flipping through Hardy, Littlewood, and Polya: The first inequality is equivalent to \[\frac{h_{k-1}(a,b,c)}{h_{k-1}(1,1,1)}\ge\frac{h_{k-2}(a,b,c)}{h_{k-2}(1,1,1)}\cdot\frac{h_1(a,b,c)}{h_1(1,1,1)}\]for $a,b,c$ not necessarily distinct, where $h_r$ denotes the $r^{th}$ complete homogeneous symmetric polynomial. (Note that $h_r(1,1,1)=\binom{r+2}{2}$.) We can prove this by using the generating function for complete homogeneous symmetric polynomials and viewing the expressions \[\frac{a^n(b-c)+b^n(c-a)+c^n(a-b)}{(a-b)(b-c)(c-a)}\]for $n\ge0$ as terms of a linear recurrence in $a,b,c$. (This also follows by Lagrange interpolation and induction on $n$.) Fortunately, there is an analog of Newton's inequalities (due to Schur): if $D=\{(x,y) | 0\le x,y,x+y\le1\}$, we can find via the generalized beta function (obtained by using two scaled beta functions) that \[\iint_{D}x^u y^v (1-x-y)^w\;dx\;dy = \frac{\Gamma(u+1)\Gamma(v+1)\Gamma(w+1)}{\Gamma(u+v+w+3)} = \frac{u!v!w!}{(u+v+w+2)!},\]so \[H_r(a,b,c)=\frac{h_r(a,b,c)}{h_r(1,1,1)}=2!\iint_{D}(ax+by+c(1-x-y))^r\;dx\;dy.\]By Cauchy-Schwarz, \[H_{r-1}(a,b,c)H_{r+1}(a,b,c)\ge H_r(a,b,c)^2,\]so finally \[\frac{H_{k-1}(a,b,c)}{H_{k-2}(a,b,c)}\ge\frac{H_{k-2}(a,b,c)}{H_{k-3}(a,b,c)}\ge\cdots\ge \frac{H_1(a,b,c)}{H_0(a,b,c)}=H_1(a,b,c),\]whence the first inequality follows. The second inequality, on the other hand, is equivalent to $H_k(a,b,c)/H_{k-2}(a,b,c)\ge (a^2+b^2+c^2)/3$. Hence it suffices to prove it for $k=4,5$, because we can then induct using the inequality $H_{r-2}H_{r+2}\ge H_r^2$ (which also follows by Cauchy). Comment: For a combinatorial interpretation of the beta function (for integer arguments), see this math.stackexchange thread.

This is the most beautiful result in mathematics I have ever seen. Well, I may not be justified to make that sarcastic remark since I don't understand the problem at all.