sorry ,but i don't find homo

tommywong

teomihai :

if you have one ideas about this:

If a,b,c>0 s .t. $a+b+c=3p $ , prove:

$a^2b+b^2c+c^2a+(p+1)(a^2+b^2+c^2)\geq{3p^2(2p+1)}$

sorry ,but i don't find homogenization.

thanks very much

Czhang271828

Prove that $a^2b+b^2c+c^2a+(p+1)(a^2+b^2+c^2)\geq{3p^2(2p+1)}$. Here $3p=a+b+c$.

*Proof.* $\Longleftrightarrow $
$$
\dfrac{1}{3}\sum_{a,b,c}(4a^2b+ab^2+c^3)+(a^2+b^2+c^2)\geq \dfrac{2(a+b+c)^3}{9}+\dfrac{(a+b+c)^2}{3}.
$$
On one hand,
$$
a^2+b^2+c^2\geq \dfrac{(a+b+c)^2}{3}.
$$
On the other hand
$$
3\sum_{a,b,c}(4a^2b+ab^2+c^3)\overset\ast \geq 2\sum_{a,b,c}(c^3+3a^2b+3ab^2+2abc)=2(a+b+c)^3.
$$
Here ($\ast$​) part is equivalent to
$$
\sum_{a,b,c}(c^3+6a^2b)\geq \sum_{a,b,c}(3ab^2+4abc).
$$
We notice that
$$
\sum_{a,b,c}(b^3+2a^2b+4a^2b)\geq \sum_{a,b,c}3ab^2+4\cdot 3abc=\sum_{a,b,c}(3ab^2+4abc).
$$
$\square$

By $\dfrac{\mathrm{ChatGPT}+\mathrm{Czhang271828}}{2}$.

hbghlyj

Czhang271828 发表于 2023-4-18 19:19
...
$$
\sum_{a,b,c}(b^3+2a^2b+4a^2b)\color{#f00}\geq \sum_{a,b,c}3ab^2+4\cdot 3abc=\sum_{a,b,c}(3ab^2+4abc).
$$

$\Large\sum_{a,b,c}(b^3+2a^2b+4a^2b)\color{#f00}\geq \sum_{a,b,c}3ab^2+4\cdot 3abc$

$\Large\impliedby \underbrace{\sum_{a,b,c}(b^3+2a^2b)\color{#f00}\geq \sum_{a,b,c}3ab^2}_\text{I don't understand😥}$   and   $\underbrace{\sum_{a,b,c}4a^2b\geq4\cdot 3abc}_\text{AM-GM}$

hbghlyj

来自@Czhang271828的证明的最后一步：
$a,b,c>0$不全相等，证明
$$\sum_\text{cyc} b (a - b) (2 a - b)>0$$
MSE

hbghlyj

$$\iff\sum_\text{cyc}\left(a(a-b)(2a-b)-\frac{1}{3}(a^3-b^3)\right)\geq0$$
居然可以配成$∑(a-b)^2(5a+b)>0$
请问这类式子的一般的配方法

hbghlyj

tommywong posted at 2023-4-18 10:06
i don't find homogenization

What is “homogenization” of this inequality?