# Problem in normalization inequalities

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?

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 $$c$$.