Homogenization, what it is in inequalities and how to utilize it to its fullest. I have just learnt about homogenization, however I don't really understand it and I can't really utilize it (https://artofproblemsolving.com/wiki/index.php/Homogenization this is where I learnt about it). Could you please explain to me how to use it optimally? Moreover I found a question on IMO 2020, which I believe can be solved using homogenization. It goes as follows:
Let the real numbers $a, b, c, d$ be such that $a\ge b\ge c \ge d >0$ and $a+b+c+d=1$. Prove that:
$(a+2b+3c+4d)a^ab^bc^cd^d<1$
I attempted to solve it by saying that RHS is equal to $(a+b+c+d)^a$, but then I realized that this doesn't help and it doesn't homogenize the inequality either. Then I tried using AM-GM on LHS and getting something which can be homogenized, however I did not succeed there either. Could you please explain to me how to use homogenization optimally and how this question could be solved using it?
 A: By AM-GM $$\prod_{cyc}a^a\leq\sum_{cyc}a^2.$$
Thus, it's enough to prove that $$(a+b+c+d)^3>(a+2b+3c+4d)(a^2+b^2+c^2+d^2),$$
which is obvious after full expanding.
A: Note that $a+2b+3c+4d \leqslant a + 3(b+c+d).$ Using the weighted AM-GM, we have
$$a^2 + b^2 + c^2 + d^2 \geqslant a^ab^bc^cd^d.$$
We need to show that
$$\left[a+3(b+c+d)\right](a^2 + b^2 + c^2 + d^2) \leqslant (a+b+c+d)^3,$$
equivalent to
$$2(b+c+d)\left[a(b+c+d)-b^2-c^2-d^2\right]+2(a+b+c+d)(bc+cd+db) \geqslant 0.$$
Which is true because $a \geqslant b \geqslant c \geqslant d.$
A: A way to work with homogeneizing an inequality is to substituite every variable $x$ by $x/R$.
This way we have $a/R+b/R+c/R+d/R=1$ and
$(a/R+2b/R+3c/R+4d/R)(a/R)^{a/R}(b/R)^{b/R}(c/R)^{c/R}(d/R)^{d/R}<1$
Or, by substituting $R=a+b+c+d$,
$(a+2b+3c+4d)/(a+b+c+d)(a/R)^{a/R}(b/R)^{b/R}(c/R)^{c/R}(d/R)^{d/R}<1$
$(a+2b+3c+4d)/(a+b+c+d) \cdot ((a/R)^{a}(b/R)^{b}(c/R)^{c}(d/R)^{d})^{1/R}<1$
$(a+2b+3c+4d)/(a+b+c+d) \cdot \left(\frac{a^ab^bc^cd^d}{R^{a+b+c+d}}\right)^{1/R}<1$
$(a+2b+3c+4d)/(a+b+c+d) \cdot \left(\frac{(a^ab^bc^cd^d)^{1/R}}{R}\right)<1$
$(a+2b+3c+4d)\cdot (a^ab^bc^cd^d)^{1/(a+b+c+d)}<(a+b+c+d)^2$
And well, man, this is ugly! But it is homogeneous as desired.
Anyway, I think there is only one possible solution for this: use
$(a^ab^bc^cd^d)^{1/(a+b+c+d)} \leq \frac{aa+bb+cc+dd}{a+b+c+d} $
and prove that
$(a+2b+3c+4d)\cdot (a^2+b^2+c^2+d^2)<(a+b+c+d)^3$.
You can use the infamous Bulldozer Technique: substitutions!
\begin{align}
a &= A+B+C+D \\
b &= B+C+D \\
c &= C+D \\
d &= D
\end{align}
where $A,B,C \geq 0,D > 0$. Expanding everything, the inequality becomes obvious. Indeed, dehomogeneizing it by putting $D=1$, we have the inequality as being the same as
$9 C^{3} + C^{2} \left(12 A + 21 B + 42\right) + C \left(A^{2} + 14 A B + 34 A + 12 B^{2} + 62 B + 60\right) + A^{2} B + 4 A B^{2} + 18 A B + 24 A + 2 B^{3} + 16 B^{2} + 44 B +  24 > 0$
In certain sense it implies the inequality is way loose.
(Yes, I used Sympy Gamma here!)
A: clean solution without any expansion
As others have noted, it is sufficient to show $$(a+b+c+d)^3 \ge (a^2+b^2+c^2+d^2)(a+2b+3c+4d)$$
write it as $$\dfrac{(a+b+c+d)^2}{a^2+b^2+c^2+d^2} \ge \dfrac{a+2b+3c+4d}{a+b+c+d}$$
subtracting 1,
$$\dfrac{2\sum_{sym} ab}{\sum a^2}\ge \dfrac{b+2c+3d}{a+b+c+d}$$
equivalent to
$$ \dfrac{2\sum_{sym} ab}{b+2c+3d}\ge \dfrac{a^2+b^2+c^2+d^2}{a+b+c+d}$$
which is true bcoz
$$ \dfrac{2\sum_{sym} ab}{b+2c+3d}\ge a \ge \dfrac{a^2+b^2+c^2+d^2}{a+b+c+d}$$
