1

If $x,y,z\ge 0: (x^2+y^2)(y^2+z^2)(z^2+x^2)=2.$ Find the maximum $$P=xy^2+yz^2+zx^2+xyz$$

I guess equality occurs when $x=y=z$ so I tried to prove homogenizing inequality $$xy^2+yz^2+zx^2+xyz \le \sqrt{2(x^2+y^2)(y^2+z^2)(z^2+x^2)}$$ Now, by well-known result $xy^2+yz^2+zx^2+xyz \le\dfrac{4}{27}(x+y+z)^3,$ we just need to prove $$\dfrac{4}{27}(x+y+z)^3\le \sqrt{2(x^2+y^2)(y^2+z^2)(z^2+x^2)} $$ It is equivalent to $$\dfrac{8}{729}(x+y+z)^6\le(x^2+y^2)(y^2+z^2)(z^2+x^2) \tag{1}$$ I am stuck to continue.

Hope you can help me prove $(1)$

Also, I am looking for other proofs. Thanks for your interest.

Michael Rozenberg
  • 203,855
  • If you substitute $y=z=x$ in the equality constraint, you end up with $8x^2=2$ which means that $x=y=z=\frac{1}{\sqrt[3]{2}}$. – JimB Aug 07 '23 at 04:29

4 Answers4

1

Another way.

Let $\{x,y,z\}=\{a,b,c\},$ where $a\geq b\geq c$.

Thus, by Rearrangement and C-S twice we obtain: $$xy^2+yz^2+zx^2+xyz=xy\cdot y+yz\cdot z+zx\cdot x+abc\leq $$ $$\leq ab\cdot a+bc\cdot c+ac\cdot b+abc=b(a+c)^2=$$ $$=\sqrt{b^2(a+c)^4}\leq\sqrt{(ab+bc)^2\cdot2(a^2+c^2)}\leq\sqrt{2(a^2+b^2)(b^2+c^2)(a^2+c^2)}=2.$$ The equality occurs for $x=y=z,$ which says that we got a maximal value.

Michael Rozenberg
  • 203,855
1

Proof.

We will prove $$xy^2+yz^2+zx^2+xyz\le \sqrt{2(x^2+y^2)(y^2+z^2)(z^2+x^2)}.\tag{*}$$ By using CBS inequality, $$xy^2+yz^2+zx^2=xy.y+yz.z+zx.x+xyz\le \sqrt{(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2)}.$$ Also by AM-GM $$xy^2+yz^2+zx^2\ge 3xyz \implies xyz+xy^2+yz^2+zx^2\le \frac{4(xy^2+yz^2+zx^2)}{3}.$$ Id est, it is enough to prove $$\frac{4}{3}\sqrt{(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2)} \le \sqrt{2(x^2+y^2)(y^2+z^2)(z^2+x^2)},$$or $$9(x^2+y^2)(y^2+z^2)(z^2+x^2)\ge 8(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2).$$ The last inequality is a well- known result which $$\iff x^2(y^2-z^2)^2+y^2(z^2-x^2)^2+z^2(x^2-y^2)^2\ge 0.$$

Hence, $(*)$ is proven. We got that maximal is equal to $2$ when $x=y=z=\frac{1}{\sqrt[3]{2}}.$

Michael Rozenberg
  • 203,855
Tran Ngoc Khuong Trang
  • 1
1

Alternative proof.

WLOG, assume that $(x-y)(z-y)\le 0.$ It implies \begin{align*} xy^2+yz^2+zx^2+xyz&=yx^2+yz^2+2xyz+x(x-y)(z-y)\\&\le yx^2+2xyz+yz^2\\&= y(x+z)^2 \end{align*} By using CBS inequality \begin{align*} y(x+z)^2&=(x+z)(yx+yz)\\&\le \sqrt{2(x^2+z^2)}.\sqrt{(x^2+y^2)(z^2+y^2)}\\&=\sqrt{2(x^2+z^2)(x^2+y^2)(z^2+y^2)}. \end{align*} The desired follows.

Tran Ngoc Khuong Trang
  • 1
0

After squaring we need to prove that: $$\sum_{cyc}(2x^4y^2+x^4z^2-2x^3y^2z-2x^3z^2y+x^2y^2z^2)\geq0,$$ which is true by AM-GM, Schur and AM-GM again: $$\sum_{cyc}(2x^4y^2+x^4z^2-2x^3y^2z-2x^3z^2y+x^2y^2z^2)\geq$$ $$\geq\sum_{cyc}(2x^3y^3+x^4y^2-2x^3y^2z-2x^3z^2y+x^2y^2z^2)=$$ $$=2\sum_{cyc}(x^3y^3-x^3y^2x-x^3z^2y+x^2y^2z^2)+\sum_{cyc}(x^4y^2-x^2y^2z^2)\geq0.$$ Thus, $P\leq2.$

The equality occurs for $x=y=z,$ which says that we got a maximal value.

Michael Rozenberg
  • 203,855