A Doubt on Stolarsky's theorem . Recently, While I was solving some problems, I saw a question which was symmetric . I thought of applying Stolarsky's inequality, but it was not homogeneous. My question is,  Can I normalize it to something, say $a+b+c=1$, put it in the inequality and apply stolarsky's inequality to get the result?. Will it be right to use Both Normalization and Stolarsky's inequality.
About the theorem, If $P(x,y,z)$ is a homogeneous polynomial of degree 3 in 3 variables, Then the following 2 statements are equivalent. 1) If $P(1,1,1)≥0,P(1,1,0)≥0,,P(1,0,0)≥0,$, Then statement 2 holds true for all positive reals 2) If the above statement is true, then $P(a,b,c)≥0$ for all positive real numbers  .
For my question, I meant, If I the the inequality $a^2+b^2+c^2+2abc+1 ⩾ 2(ab+bc+ca)$ Can I consider $a+b+c=1$ And make the polynomial homogeneous to apply Stolarsky's inequality.
 A: First, if you want to show $a^2+b^2+c^2+2abc \ge 2(ab+bc+ca)$ holds for all positive reals, note that you cannot "homogenize" it by assuming a constraint like $a+b+c=1$.  This would only prove the inequality for a subset of positive reals.
As an e.g., suppose I want to show that $a+b+c \le 1$ for all positive reals (which obviously is false). However if I homogenise it using your assumption, I get a true statement, $a+b+c \le a+b+c$.  So clearly that isn't a valid method.

As an unrelated matter, in your particular case, the inequality can be proven by using the symmetry - so some two variables, WLOG say $a, b$ are on the same side as $1$, and hence $2(a-1)(b-1)c+(c-1)^2+(a-b)^2 \ge 0$

OTOH if you are adamant about using the theorem you quoted to prove this, consider $a+b+c=k$ and use that to homogenise.  If you can prove that for any $k \ge 0$, the inequality holds, you are through! So in this approach your homogeneous version is
$$P(a, b, c): k^2(a^2+b^2+c^2)(a+b+c) + 2k^3abc + (a+b+c)^3 - 2k^2(a+b+c)(ab+bc+ca) \ge 0$$
now $P(1, 1, 1) = 2k^3+27-9k^2 \ge 3\sqrt[3]{27k^6}-9k^2 \ge 0$ by AM-GM,
$P(1, 1, 0) = 4k^2+8-4k^2 > 0$ and $P(1, 0, 0) = k^2+1> 0$ hence you can use the theorem to say for any $k \ge 0$ we must have $P \ge 0$ so we have our proof.  
Maybe there are inequalities which become simpler using obscure theorems, but personally I would go for the more well known theorems for solutions. 
A: Your question is extremely vague. There's a generalization of Stolarsky's inequality to arbitrary weights in this paper: Stolarsky Inequality with general weights. In case you don't have access to the above the theorem is:
Let $w_i$ be integrable weights which are nonnegative for $i=1,2,3$ defining:
$$W_i(x):=\frac{\int_0^xw_i(t)dt}{\int_0^1 w_i(t)dt},$$
and define:
$$Q(g,w)=\frac{\int_0^1 g(w)w(x)dx}{\int_0^1w(x)dx}.$$
If $g$ is a function of bounded variation on $[0,1]$ such that $0\leq g(1)\leq g(x)\leq g(0)$ for all $x\in (0,1)$ and if
$$W_1(x)W_2(x)=W_3(x), \mbox{ for all }x\in[0,1],$$
then
$$g(0)Q(g,w_3)\geq Q(g,w_1)Q(g,w_2).$$
