Problem in normalization inequalities

Question

Let $a,b,c>0$. Prove that: $$\dfrac{(a+b)^2(b+c)^2(c+a)^2}{abc} \ge \dfrac{64}{27}(a+b+c)^3$$ First solution:

$\bullet$ Since the inequality is homogeneous, we may normalize $a+b+c=3$, we need to prove: $$(a+b)(b+c)(c+a)\ge8\sqrt{abc}$$

$\bullet$ Notice that, $a+b+c=3$, we may have $$(a+b)(b+c)(c+a)=(3-a)(3-b)(3-c)=3(ab+bc+ca)-abc \ge 3.\sqrt{3abc(a+b+c)}-abc=9\sqrt{abc}-abc$$

$\bullet$ So we need to prove $$9\sqrt{abc}-abc \ge 8\sqrt{abc}$$ or $$abc\le1$$

which is true by AM-GM: $$abc\le\dfrac{(a+b+c)^3}{27}=1$$

Second solution:

$\bullet$ Since the inequality is homogeneous, we may normalize $a+b+c=1$, we need to prove: $$27(a+b)^2(b+c)^2(c+a)^2\ge64abc$$

$\bullet$ This solution is more convenient, by two famous inequalities $(a+b)(b+c)(c+a)\ge\dfrac{8}{9}(a+b+c)(ab+bc+ca)$ and $(ab+bc+ca)^2\ge3abc(a+b+c)$, the problem is solved

My question:

1. We can normalize $a+b+c=k$ with any number $k$, but why the author find $k=3$ or $k=1$ and make the problem very simple.
2. I have seen in many proofs for inequalities, the author normalize $abc=k;(a+b)(b+c)(c+a)=k;ab+bc+ca=k$,...Can you explain to me how to know which expression we need to normalize? Can you start from this post with a different normalization method, or can you give me some other examples?
3. Is there a way to solve this problem without using the normalization method?

Answer 1

3. Yes, of course.

For example, after full expanding we need to prove that: $$\sum_{sym}(27a^4b^2-5a^4bc+27a^3b^3-30a^3b^2c-19a^2b^2c^2)\geq0,$$ which is true by Muirhead.

Also, we can prove that the inequality $$(a+b)^2(a+c)^2(b+c)^2\geq\frac{64}{27}abc(a+b+c)^3$$ is true for any reals $a$, $b$ and $c$.