How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?

Question

How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?

Answer 1

You can write the inequality as $$ xyz(x+y+z)\leq\frac13. $$ On the left-hand-side, using Cauchy-Schwarz, $$ xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz. $$ Now we get the seemingly easier problem of maximizing $xyz$ under $x^2+y^2+z^2=1$. Here symmetry, common sense, or Lagrange multipliers show that the maximum is achieved when $x=y=z$. Then $$ xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz\leq\frac{\sqrt3}{(\sqrt3)^3}=\frac13. $$

Answer 2

By Cauchy–Schwarz inequality with

• $u=(x,y,z)$
• $v=(xyz,xyz,xyz)$

we have

$$u\cdot v\le |u|\cdot |v|$$

that is

$$x^2yz+xy^2z+xyz^2 \leq\sqrt3\,xyz \le \frac{1}{3}$$

indeed by AM-GM

$$\frac13=\frac{x^2+y^2+z^2}{3}\ge \sqrt[3]{x^2y^2z^2}\implies xyz \le\frac{1}{3\sqrt3}$$

Answer 3

It's true even for all reals $x$, $y$ and $z$.

Indeed, we need to prove that $$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ or $$\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\geq0$$ or $$\sum_{cyc}(2x^4+4x^2y^2-6x^2yz)\geq0$$ or $$\sum_{cyc}(x^4-2x^2y^2+y^4+3z^2x^2-6z^2xy+3z^2y^2)\geq0$$ or $$\sum_{cyc}(x-y)^2((x+y)^2+3z^2)\geq0,$$ which is obvious.