An algebra problem by Ravneet Singh

Algebra Level 4

$\Large xy + yz + zx$

If $x, y, z$ are positive real numbers and $xyz = 2017$ , which of the given options is not a possible value of above expression?

951

4034

2017

478

1 solution

Md Zuhair
Jun 2, 2017

$\color{#3D99F6}\text{By AM-GM Inequality}$

We have,

$\dfrac{xy+yz+zx}{3} \geq \sqrt[3]{x^2y^2z^2}$

$xy+yz+zx \geq 3\sqrt[3]{x^2y^2z^2}$

Now $xyz=2017$

So putting the value,

We get

$xy+yz+zx \geq 3\sqrt[3]{x^2y^2z^2} \implies xy+yz+zx \geq 478.915$

So $xy+yz+zx \geq \boxed{478.915}$

So $478$ is not possible answer.

This solution is incomplete. Do you see why?

Calvin Lin Staff - 4 years ago

The solution shows what the minimum of the expression can be. It does not give any info about its maximum.

Tapas Mazumdar - 4 years ago

Precisely! In particular, it does not explain that the rest of the numbers can be achieved.

Calvin Lin Staff - 4 years ago

@Calvin Lin – But I am not able to find any maxima for this. Can you help me out? Is it $\infty$ ?

Tapas Mazumdar - 4 years ago

@Tapas Mazumdar – Same, I am too having difficulty to find it out.

Md Zuhair - 4 years ago

@Md Zuhair – Can you achieve (approximately) $2017^2? 2017^3$ ? How can we show that "all large enough values can be achieved"?

Calvin Lin Staff - 4 years ago

@Calvin Lin – This is what I have in my mind. Since there is no restriction on distinct values for the variables we can fix $x=y$ , now we see that

$x^2 z = 2017 \implies z = \dfrac{2017}{x^2}$

Letting $x \to 0$ gives $z \to \infty$ and so the expression

$xy + yz + zx = 2017 \left( \dfrac 1x + \dfrac 1y + \dfrac 1z \right) = 2017 \left( \dfrac 2x + \dfrac 1z \right)$

which also tends to $\infty$ as $x \to 0$ and $z \to \infty$ .

Tapas Mazumdar - 4 years ago

@Tapas Mazumdar – Great! So now we've established that any number that is "large enough" can be achieved.

How much is "large enough" though? IE can every number between 478 and "large enough" be achieved?

Calvin Lin Staff - 4 years ago