Nice inequality problem!

Upvote if you satisfied

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Applying $AM$ $\geq$ $GM$

$\dfrac{x+y+z} {3} \geq \sqrt[3]{xyz}$

$\implies$ $\dfrac{6}{3} \geq \sqrt[3]{xyz}$

$\implies$ $xyz\leq 2^3$

So maximum value of $\boxed {xyz = 8}$

$\large \color{#D61F06} 2 \times 2 \times 2 = 8, \text{ therefore xyz = 8.}$

I agree with Pi to make this question valid you should specify positive numbers.only.

X = any negative number say x = -1,000,000

Y= - X +7 y= 1,000,007

z =-1

there is no limit how large xyz can be.