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Problem:

Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$


I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$.

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    $\begingroup$ use the Lagrange Multiplier methode $\endgroup$ Commented Jan 9, 2016 at 22:47
  • $\begingroup$ I get $\sqrt{\frac{11}{2}}$ at $\left(\sqrt{\frac{3}{22}},\sqrt{\frac{11}{22}},\sqrt{\frac{8}{22}}\right)$. $\endgroup$
    – CommonerG
    Commented Jan 9, 2016 at 23:20

3 Answers 3

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Let $a,b\in\mathbb{R}$. We shall maximize $axy+byz$ subject to $x^2+y^2+z^2=1$ ($x,y,z\in\mathbb{R}$).

By Cauchy-Schwarz, $axy+byz=(ax+bz)y\leq \left(\sqrt{a^2+b^2}\sqrt{x^2+z^2}\right)|y|$. By AM-GM, $\sqrt{x^2+z^2}\,|y|\leq \frac{\left(x^2+z^2\right)+y^2}{2}=\frac{1}{2}$. That is, $axy+byz\leq \frac{\sqrt{a^2+b^2}}{2}$. The equality holds iff $\left(x,y,z\right)=\pm\left(\frac{a}{\sqrt{2\left(a^2+b^2\right)}},\frac{1}{\sqrt{2}},\frac{b}{\sqrt{2\left(a^2+b^2\right)}}\right)$.

(Similarly, we also have the lower bound $axy+byz\geq -\frac{\sqrt{a^2+b^2}}{2}$. The equality holds iff $\left(x,y,z\right)=\pm\left(\frac{a}{\sqrt{2\left(a^2+b^2\right)}},-\frac{1}{\sqrt{2}},\frac{b}{\sqrt{2\left(a^2+b^2\right)}}\right)$.)

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  • $\begingroup$ Pretty neat. Thank you. $\endgroup$ Commented Jan 9, 2016 at 23:03
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    $\begingroup$ I think your $y-$ and $z-$coördinates should be reversed in your solutions. $\endgroup$
    – Théophile
    Commented Jan 10, 2016 at 0:15
  • $\begingroup$ @Théophile Thanks, fixed. $\endgroup$ Commented Jan 10, 2016 at 0:52
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We can use AM-GM inequality alone to settle the question. To this end, we find the constants $A,B,C,D$ such that: $\sqrt{6}xy \leq A^2x^2+B^2y^2$, and $4yz \leq C^2y^2+D^2z^2$ and $A^2 = B^2+C^2 = D^2, AB = \dfrac{\sqrt{6}}{2}, CD = 2 = AC$. Thus $A^4 = A^2\cdot A^2 = A^2(B^2+C^2) = (AB)^2 + (AC)^2 = \left(\dfrac{\sqrt{6}}{2}\right)^2 + 2^2 = \dfrac{11}{2}=\dfrac{22}{4}$. So $A^2 = \dfrac{\sqrt{22}}{2}$, and clearly this value of $A^2 = \dfrac{\sqrt{22}}{2}$ is the maximum we are looking for.

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The easiest method seems to be to use Lagrange multipliers. $$ L = \sqrt{6} xy + 4 yz - \frac{1}{2} \lambda (x^2 + y^2 + z^2 - 1) \, , $$ for some $\lambda$. Looking at derivatives $$ \frac{\mathrm{d} L}{\mathrm{d} x} = \sqrt{6} y - \lambda x = 0 $$ $$ \frac{\mathrm{d} L}{\mathrm{d} y} = \sqrt{6} x + 4z - \lambda y = 0 $$ $$ \frac{\mathrm{d} L}{\mathrm{d} z} = 4y - \lambda z = 0 $$ Another equation is the original constraint $x^2 + y^2 + z^2 = 1$. Now one has to solve the set of equations. After a few steps one gets $\lambda^2 = 22$ and $x = \pm \sqrt{3/22}$. From that it follows that $y = \pm \sqrt{1/2}$ and $z = \pm \sqrt{4/11}$.

The signs of $x,y,z$ are not independent, but the maximum of $\sqrt{6}xy + 4 yz$ is at $(\sqrt{3/22},\sqrt{1/2},\sqrt{4/11})$ and it is $\sqrt{11/2}$.

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  • $\begingroup$ Thanks for introducing me to Lagrange's multipliers. I didn't know about it until now, and it certainly seems very helpful for questions such as this one. $\endgroup$ Commented Jan 13, 2016 at 15:26

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