11
$\begingroup$

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

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

AND

$$xy + yz + zx = 3$$

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.

$\endgroup$
  • $\begingroup$ 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
8
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.