Let $x, y$ and $z$ be three real numbers satisfying the following conditions:

$$0 < x \leq y \leq z$$


$$xy + yz + zx = 3$$

Prove that the maximum value of $(x y^3 z^2)$ is $2.$

I tried using the weighted AM-GM inequality, but to no avail as the powers 1,2 and 3 are giving me a hard time. How should I proceed? Thanks in advance.

  • $\begingroup$ have you tried the Lagrange Multiplier method? $\endgroup$ – Dr. Sonnhard Graubner Feb 24 '17 at 10:47
  • $\begingroup$ I'm not aware of that. I'll look it up. But I'm sure there's another way isn't there? $\endgroup$ – Newton Feb 24 '17 at 10:47
  • $\begingroup$ @Dr.SonnhardGraubner. How would you take into account $0 < x \leq y \leq z$ using Lagrange multipliers ? $\endgroup$ – Claude Leibovici Feb 24 '17 at 10:59

Let $x=\frac{a}{2\sqrt2}$, $y=\sqrt2b$ and $z=\sqrt2c$.

Hence, $c\geq b$ and by AM-GM: $$6=4bc+ab+ac\geq6\sqrt[6]{(bc)^4(ab)(ac)}=6\sqrt[6]{a^2b^5c^5}\geq6\sqrt{a^2b^6c^4},$$ which gives $$1\geq ab^3c^2=\frac{1}{2}xy^3z^2.$$ The equality occurs for $x=\frac{1}{2\sqrt2}$ and $y=z=\sqrt2$

and we are done!

  • $\begingroup$ Nice answer, but may I know how you guessed the substitutions for a, b and c? The numbers seem pretty random to me. $\endgroup$ – Newton Feb 24 '17 at 13:29
  • 3
    $\begingroup$ @Newton Firstly I found an equality case and from this I built the proof. $\endgroup$ – Michael Rozenberg Feb 24 '17 at 14:46

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