Find $\max(x^2y+y^2z+z^2x+xyz)(x^2z+y^2x+z^2y+xyz)$ subject to $x+2y+3z=4$

Question

Let $x$, $y$ and $z$ be non-negative numbers such that $x+2y+3z=4.$ Find: $$\max(x^2y+y^2z+z^2x+xyz)(x^2z+y^2x+z^2y+xyz).$$

I took this problem here: https://dxdy.ru/topic18767-30.html a last post.

This problem is a similar to many contests problems.

On one of Canadians olimpiads was $x^2y+y^2z+z^2x\leq4$ for non-negatives $x$, $y$ and $z$ such that $x+y+z=3.$

Also, $x^2y+y^2z+z^2x+xyz\leq4$ with the same conditions.

My attempts:

For $(x,y,z)=(2,1,0)$ we get a value $8$, which looks as a maximal value.

I solved this problem for $x=\min\{x,y,z\}$ and for $y=\min\{x,y,z\}$.

But for $z=\min\{x,y,z\}$ we need to prove that $$(x+2y+3z)^6\geq512(x^2y+y^2z+z^2x+xyz)(x^2z+y^2x+z^2y+xyz),$$ which after substitution $x=z+u$, $y=z+v$ gives something very hard: $$38464z^6+64(473u+1202v)z^5+16(447u^2+3068uv+4092v^2)z^4+$$ $$+32(7u^3+234u^2v+1044uv^2+952v^3)z^3+$$ $$+4(7u^4-200u^3v+808u^2v^2+3040uv^3+2032v^4)z^2+$$ $$+4(9u^5-38y^4v-152u^3v^2+208u^2v^3+592uv^4+288v^5)z+$$ $$+(u-2v)^2(u^4+16u^3v+120u^2v^2+64uv^3+16v^4)\geq0.$$ Thank you!

Answer 1

Solution by Tran Quoc Anh.

We need to prove $$8\left(a^{2} b+b^{2} c+c^{2} a+a b c\right)\left(a b^{2}+b c^{2}+c a^{2}+a b c\right) \leqslant \left(\frac{a+2 b+3 c}{2}\right)^{6} .$$ By the AM-GM inequality we have $$ \text{LHS} \leqslant \left[a^{2} b+b^{2} c+c^{2} a+a b c+2\left(a b^{2}+b c^{2}+c a^{2}+a b c\right)\right]^{2}.$$ and $$(a+2 b)(b+2 c)(c+2 a)=2\left[\left(a^{2} b+b^{2} c+c^{2} a\right)+2\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c\right]+3 a b c$$ $$ \geqslant 2\left[\left(a^{2} b+b^{2} c+c^{2} a\right)+2\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c\right],$$ and $$\begin{aligned} (a+2 b)(b+2 c)(c+2 a) &=\frac{1}{4} \cdot(a+2 b) \cdot 4(b+2 c) \cdot(c+2 a) \\ & \leq \frac{1}{4}\left[\frac{(a+2 b)+4(b+2 c)+(c+2 a)}{3}\right]^{3} \\ &=\frac{1}{4}(a+2 b+3 c)^{3}.\end{aligned}$$ Therefore $$\left(a^{2} b+b^{2} c+c^{2} a\right)+2\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c \leq \frac{1}{8}(a+2 b+3 c)^{3}.$$ The proof is completed.

Answer 2

Proceeding along Michael Rozenberg's approach, it suffices to prove that \begin{align} &32(7u^3+234u^2v+1044uv^2+952v^3)z^3\\ & + 4(7u^4-200u^3v+808u^2v^2+3040uv^3+2032v^4)z^2\\ & + 4(9u^5-38u^4v-152u^3v^2+208u^2v^3+592uv^4+288v^5)z\\ &+ (u-2v)^2(u^4+16u^3v+120u^2v^2+64uv^3+16v^4)\ge 0. \end{align} If $v=0$, clearly the inequality is true. If $v>0$, due to homogeneity, assume that $v=1$. It suffices to prove that \begin{align} &32(7u^3+234u^2+1044u+952)z^3\\ & + 4(7u^4-200u^3+808u^2+3040u+2032)z^2\\ & + 4(9u^5-38u^4-152u^3+208u^2+592u+288)z\\ &+ (u-2)^2(u^4+16u^3+120u^2+64u+16)\ge 0. \end{align} There are two possible cases:

1) $z \ge 1$: Let $z = 1 + w$ for $w\ge 0$. It suffices to prove that $$a_3w^3 + a_2w^2 + a_1w + a_0\ge 0$$ where \begin{align} a_3 &= 224 u^3+7488 u^2+33408 u+30464, \\ a_2 &= 28 u^4-128 u^3+25696 u^2+112384 u+99520, \\ a_1 &= 36 u^5-96 u^4-1536 u^3+29760 u^2+126912 u+108800, \\ a_0 &= u^6+48 u^5-64 u^4-1536 u^3+11792 u^2+48128 u+39808. \end{align} It is easy to prove that $a_3, a_2, a_1, a_0\ge 0$ for $u\ge 0$. True.

2) $z \in [0, 1)$: Let $z = \frac{s}{1+s}$ for $s\ge 0$. It suffices to prove that $$b_3s^3 + b_2s^2 + b_1s + b_0 \ge 0$$ where \begin{align} b_3 &= u^6+48 u^5-64 u^4-1536 u^3+11792 u^2+48128 u+39808, \\ b_2 &= 3 u^6+108 u^5-96 u^4-3072 u^3+5616 u^2+17472 u+10624, \\ b_1 &= 3 u^6+72 u^5+28 u^4-1664 u^3+1552 u^2+2944 u+1344, \\ b_0 &= u^6+12 u^5+60 u^4-352 u^3+240 u^2+192 u+64. \end{align} It is easy to prove that $b_3, b_2, b_1, b_0\ge 0$ for $u\ge 0$. True.

We are done.

Answer 3

Here is my second proof.

By AM-GM, it suffices to prove that $$(x^2y + y^2z + z^2x + xyz) + 2(xy^2 + yz^2 + zx^2 + xyz) \le 8.$$ (Note: When $(x, y, z) = (2, 1, 0)$, we have $x^2y + y^2z + z^2x + xyz= 4$ and $xy^2 + yz^2 + zx^2 + xyz=2$. This motivates the above form of AM-GM.)

It suffices to prove that, for all $x, y, z \ge 0$, $$(x^2y + y^2z + z^2x + xyz) + 2(xy^2 + yz^2 + zx^2 + xyz) \le 8\cdot \frac{(x + 2y + 3z)^3}{64}. \tag{1}$$

If $z = 0$, we have $$\mathrm{RHS}_{(1)} - \mathrm{LHS}_{(1)}= \frac18(x + 2y)^3 - 2xy^2 - x^2 y = \frac18(x + 2y)(x - 2y)^2 \ge 0.$$

If $z > 0$, WLOG, assume that $z = 1$. It suffices to prove that $$(x^2y + y^2 + x + xy) + 2(xy^2 + y + x^2 + xy) \le 8\cdot \frac{(x + 2y + 3)^3}{64}$$ or \begin{align*} &y^3 + \left(-\frac12 x + \frac72\right)y^2 + \left(-\frac14x^2 + \frac32 x + \frac{19}{4}\right) y\\[6pt] &\qquad+ \frac18x^3 - \frac78x^2 + \frac{19}{8}x + \frac{27}{8} \ge 0. \tag{2} \end{align*}

Using $(y - x/2)^2 y \ge 0$, we have $y^3 \ge xy^2 - \frac14x^2y$. It suffices to prove that \begin{align*} &\left(xy^2 - \frac14x^2y\right) + \left(-\frac12 x + \frac72\right)y^2 + \left(-\frac14x^2 + \frac32 x + \frac{19}{4}\right) y\\[6pt] &\qquad + \frac18x^3 - \frac78x^2 + \frac{19}{8}x + \frac{27}{8} \ge 0 \end{align*} or $$\left(\frac12 x + \frac72\right)y^2 + \left(-\frac12x^2 + \frac32 x + \frac{19}{4}\right) y + \frac18x^3 - \frac78x^2 + \frac{19}{8}x + \frac{27}{8} \ge 0. \tag{3}$$ which is true since \begin{align*} &4\cdot \left(\frac12 x + \frac72\right)\cdot \left(\frac18x^3 - \frac78x^2 + \frac{19}{8}x + \frac{27}{8}\right) - \left(-\frac12x^2 + \frac32 x + \frac{19}{4}\right)^2\\[6pt] ={}& \frac32x^3 - 5x^2 + \frac{103}{4} x + \frac{395}{16}\\[6pt] >{}& 0. \end{align*}

We are done.