5

Prove that $\displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right)$, for $x$, $y$, $z\in\Bbb R_+$.

The $pqr$ method doesn't seem possible because the power is too high. $$\iff\left(p^2-2q\right)^2\ge9r\left(p^3-3pq+3r\right).$$

Then expand the expression to get $$\sum x^6+3\sum\left(x^4y^2+x^2y^4\right)+6x^2y^2z^2\ge9\sum x^4yz.$$ I wanted to use SOS but cannot find the weight of three squares, my progress: $$3\sum x^4(y-z)^2=3\sum\left(x^4y^2+x^2y^4\right)-6\sum x^4yz.$$ Whats left is $\displaystyle\sum x^6+6x^2y^2z^2-3\sum x^4yz$. I have trouble dealing with it.

Michael Rozenberg
  • 194,933
  • I'm curious about the way of proving that stronger inequality you obtained by SOS , $\sum x^6+6x^2y^2z^2\geqslant3\sum x^4yz$ . I tried to divide both sides by $(xyz)^2$ and let $a=(x^2)/(yz),~\ldots$ , which yields $\sum a^2+6\geqslant3\sum a$ for $abc=1$ . That's the furthest I can go. – O-17 Jan 19 '23 at 19:10

6 Answers6

2

Update: A simpler pqr

It suffices to prove that $$\left(p^2-2q\right)^3\ge9r\left(p^3-3pq+3r\right).$$

Using $q^2 \ge 3pr$ and $pq \ge 9r$, it suffices to prove that $$\left(p^2-2q\right)^3\ge 9 \cdot \frac{q^2}{3p}\cdot\left(p^3-3pq+3 \cdot \frac{pq}{9}\right)$$ or $$(p^2 - 2q)^3 \ge q^2(3p^2 - 8q)$$ or $$(p^2 - 2q)^3 + (2q)^3 \ge 3p^2 q^2$$ or (using $x^3 + y^3 = (x + y)[(x + y)^2 - 3xy]$) $$p^2[p^4 - 3(p^2 - 2q)\cdot 2q] \ge 3p^2 q^2$$ or $$p^2 (p^2 - 3q)^2 \ge 0$$ which is true.

We are done.

River Li
  • 37,323
  • Good. But what's the idea for transforming the inequality about $p$, $q$ and $r$? When I'm doing this, it's like trying my luck. –  Jan 18 '23 at 15:40
  • @youthdoo There are some most used relations among $p, q, r$. In experience, $q^2 \ge 3qr$, $p^2 \ge 3q$ and degree 3/4 Schur's inequalities etc. are often used. If all of them fail, you may used the ultimate $$0 \le (a - b)^2(b - c)^2(c - a)^2 = - 4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2$$ which is quadratic in $r$ (gives you $\cdots \le r \le \cdots$). – River Li Jan 19 '23 at 00:06
  • Thanks, so is it that\[0 \le (a - b)^2(b - c)^2(c - a)^2 = - 4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2.\]is the necessary sufficient condition right? –  Jan 19 '23 at 00:09
  • @youthdoo Yes, $a, b, c\ge 0$ is equivalent to $p, q, r \ge 0$ and $- 4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2\ge 0$ (the cubic $u^3 - pu^2 + qu -r = 0$ has three real non-negative roots). – River Li Jan 19 '23 at 00:15
  • @youthdoo You may see my answer pqr for the book listing often used pqr relations. – River Li Jan 19 '23 at 00:27
2

This is a proof that continues your SOS attempt, which is to prove

$$ \sum x^6+6x^2y^2z^2\geqslant3\sum x^4yz $$

By noticing the following identity, this inequality is obviously true by Schur's inequality

$$ \sum(x^6+6x^2y^2z^2-3x^4yz)=\sum x^4(x-y)(x-z)+4(x-y)^2(y-z)^2(z-x)^2+2\sum yz(x-y)^2(x-z)^2+2xyz\sum x(x-y)(x-z) $$

This proof is by brute force using the Triangle of Coefficient Method. Though I've found a way to simplify this inequality, I didn't managed to solve it that way.

Namely, by dividing both sides with $(xyz)^2$ , we obtain

$$ \sum\left(\frac{x^2}{yz}\right)^2+6\geqslant3\sum\frac{x^2}{yz} $$

Let $a=(x^2)/(yz),~b=(y^2)/(zx),~c=(z^2)/(xy)$ , we have to prove that

$$ a^2+b^2+c^2+6\geqslant 3(a+b+c) $$

for positive real numbers $a,b,c$ such that $abc=1$ .

I hope someone would continue my attempt.

Edit. This edition contains a new proof of this inequality using SOS Method by LasterCircle

$$ \text{LHS}-\text{RHS}=\frac12\sum\left[\sum x^4+2xy(x^2+xy+y^2)-2xyz\sum x\right](x-y)^2 $$

It's apparent that $\sum x^4+2xy(x^2+xy+y^2)-2xyz\sum x>0$ , the inequality is true.

Edit2. River Li has given a nice proof in the comment.

And here's a simpler SOS

$$ \sum(x^6-3x^4yz)+6x^2y^2z^2=\frac1{18}\sum\left(x(x^2-yz)\right)^2+\frac12\sum x^2(x^2-y^2)(x^2+y^2)+\frac12\sum{x^2y^2(x-y)^2}\geqslant0 $$

O-17
  • 438
  • Do you mean Chinese Dumbass Notation? –  Jan 20 '23 at 00:15
  • @youthdoo Right, that's its another name (probably more widely known) , I have made some progress on this inequality and I'll show it in the next edition. – O-17 Jan 20 '23 at 05:04
  • But isn't the last inequality obvious. Because $a^2-a\ge-\dfrac14$? –  Jan 20 '23 at 05:10
  • @youthdoo Please note that $\sum a^2\ne\left(\sum a\right)^2$ . – O-17 Jan 20 '23 at 05:20
  • @O-17 Some typos? $a^2+b^2+c^2+6\ge 3(a+b+c)$? – River Li Jan 20 '23 at 12:42
  • 1
    @O-17 By the way, a proof for the above: WLOG, assume that $(b-1)(c-1)\ge 0$, we have$a^2+b^2+c^2+6-3(a+b+c) + (abc-1) $ $$= \frac12(a + b + c- 3)^2
    • \frac12(a - 1)^2 + a(b - 1)(c - 1) + \frac12 (b-c)^2 \ge 0.$$
     – River Li Jan 20 '23 at 13:03
  • 1
    @RiverLi Thanks for correcting, and your proof is very nice too. – O-17 Jan 20 '23 at 14:22
  • Your method is called Schur splitting, which seldom works. But it miraculously solved this inequality... –  Jan 31 '23 at 11:42
  • @youthdoo I don't know if my method is Schur splitting, but I definitely wouldn't call this method 'seldom works'. I acknowledged from your homepage that you're also from China, welcome to 'the seminar of Triangle of Coefficient' (q gruop: 875413273) – O-17 Jan 31 '23 at 14:09
  • @youthdoo By the way, your inequality may be refined to $\sum a^6+(24-9\sqrt{3})\sum a^2b^2c^2\geqslant(9-3\sqrt{3})\sum a^4bc$ by a theorem that is introduced in a book in that q group. It may be proved by its equivalence to $\sum\left((a^2+\sqrt{3}bc)\sum\left((a+b-\sqrt{3}c)(a-b)^2\right)\right)\geqslant0$ , which is obviously true. – O-17 Jan 31 '23 at 14:19
  • The problem is, we need every single coefficient non-negative to conclude a polynomial $\ge0$, and very often one of them $<0$, and we need to do further work. –  Jan 31 '23 at 15:16
  • Which question? Proving $\sum x^6+6x^2y^2z^2\ge3\sum x^4yz$? –  Feb 01 '23 at 02:26
1

Since the expression is symmetric, WLOG, $ z \geq y \geq x$.
Let $ x = a, y = a + b, z = a+b+c$, where $a, b, c \geq 0$.

Then, expanding out $(x^2 + y^2 + z^2)^3 - 9 xyz(x^3+y^3+z^3)$ gives us, courtesy of Wolfram:

$9 a^4 b^2 + 9 a^4 b c + 9 a^4 c^2 + 28 a^3 b^3 + 42 a^3 b^2 c + 30 a^3 b c^2 + 8 a^3 c^3 + 42 a^2 b^4 + 84 a^2 b^3 c + 78 a^2 b^2 c^2 + 36 a^2 b c^3 + 12 a^2 c^4 + 30 a b^5 + 75 a b^4 c + 90 a b^3 c^2 + 60 a b^2 c^3 + 27 a b c^4 + 6 a c^5 + 8 b^6 + 24 b^5 c + 36 b^4 c^2 + 32 b^3 c^3 + 18 b^2 c^4 + 6 b c^5 + c^6.$

Since all the coefficients are non-negative, this is clearly non-zero, hence we are done.

Calvin Lin
  • 68,864
  • Is this method called the Buffalos Way? –  Jan 19 '23 at 00:52
  • @youthdoo Yes. Often, the more common way BW appears is to set $ x = \min{x, y, z}, y = x+u, z = x+v$, and still requiring a bunch of SOS to resolve. In this case, this approach immediately yields non-negative coefficients, so we are done. – Calvin Lin Jan 20 '23 at 14:14
  • A bit irrelevant but how do you copy the result from wolfram alpha to paste here? Use ocr? –  Jan 21 '23 at 05:38
  • Hover on the relevant line, click on the "Plain text" tab, then click the "Copyable Plain text" which will copy it to clipboard, then paste it. – Calvin Lin Jan 21 '23 at 14:44
1

This inequality is true for any real variables.

Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ may be negative and $xyz=w^3$.

Thus, we need to prove that: $$(9u^2-6v^2)^3\geq9w^3(27u^3-27uv^2+3w^3)$$ or $f(w^3)\geq0,$ where $f$ is a concave function.

But, a concave function gets a minimal value for an extremal value of $w^3$, which by uvw happens for equality case of two variables.

Since our inequality is symmetric, even degree and homogeneous, it's enough to assume that $y=z=1$ (for $y=z=0$ the equality is obvious), which gives $$(x^2+2)^3-9x(x^3+2)\geq0$$ or $$(x-1)^2(x^4+2x^3-2x+8)\geq0,$$ which is true because $$x^4+2x^3-2x+8=(x^2+x-1)^2+x^2+7>0.$$

Michael Rozenberg
  • 194,933
1

Another way.

By AM-GM we obtain: $$(x^2+y^2+z^2)^3-9xyz(x^3+y^3+z^3)=$$ $$=\sum_{cyc}(x^6+3x^4y^2+3x^4z^2+2x^2y^2z^2-9x^4yz)=$$ $$=\frac{1}{2}\sum_{cyc}(2x^6-x^4y^2-x^4z^2+7x^4y^2+7x^4z^2-14x^4yz-4x^4yz+4x^2y^2z^2)=$$ $$=\frac{1}{2}\sum_{cyc}(x-y)^2((x^2+y^2)(x+y)^2+7z^4-2xyz(x+y+z))=$$ $$=\frac{1}{2}\sum_{cyc}(x-y)^2((z^2-xy)^2+6z^4-2xy(x+y)z+(x^2+y^2)(x+y)^2-x^2y^2)\geq$$ $$\geq\frac{1}{2}\sum_{cyc}(x-y)^2\left(6z^4-\frac{1}{2}(x+y)^3z+\frac{7}{16}(x+y)^4\right)=$$ $$=\frac{1}{2}\sum_{cyc}(x-y)^2\left(6z^4+3\cdot\frac{7}{48}(x+y)^4-\frac{1}{2}(x+y)^3z\right)\geq$$ $$\geq\frac{1}{2}\sum_{cyc}(x-y)^2\left(4\sqrt[4]{6z^4\left(\frac{7}{48}(x+y)^4\right)^3}-\frac{1}{2}(x+y)^3z\right)=$$ $$=\frac{1}{2}\sum_{cyc}(x-y)^2(x+y)^3z\left(4\sqrt[4]{6\left(\frac{7}{48}\right)^3}-\frac{1}{2}\right)\geq0.$$

Michael Rozenberg
  • 194,933
0

Hint :

Due to homogeneization we have :

$$\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)$$

We need to show for $x\ge 1$ and WLOG $c\geq 1$

$$f(x)=\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)\geq 0$$

The function is convex on $(0,\infty)$ because we have :

$$f''(x)=6(c^{6}+6c^{3}x^{2}+2c^{3}-18cx^{2}+5x^{4}+6x^{2}+1)>0$$

So we use the tangent line method we have :

$$f(x)\geq f(1)(x-1)+f(1)=(6 (c^3 + 2)^2 - 9 c (c^3 + 2) - 27 c)(x-1)+(c^3 + 2)^3 - 9 c (c^3 + 2)\geq 0$$

Because :

$$(6(c^{3}+2)^{2}-9c(c^{3}+2)-27c)\geq 0,(c^3 + 2)^3 - 9 c (c^3 + 2)=(c^3 + 2) (c - 1) (c^5 + c^4 + c^3 + 5 c^2 + 5 c - 4)\geq 0$$

Miss and Mister cassoulet char
  • 3,511