Let x, y, z be three non-negative integers su | vedclass

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Question

(c)

We have, $x+y+z=10$

Let three number $x+1, y+1, z+1$

$AM \geq GM$

$\quad \frac{(x+1)+(y+1)+(z+1)}{3} \geq [(x+1)(y+1)(z+1)]^{1 /

Vedclass · Accepted Answer

$69$

Vedclass · Answer

(c)

We have, $x+y+z=10$

Let three number $x+1, y+1, z+1$

$AM \geq GM$

$\quad \frac{(x+1)+(y+1)+(z+1)}{3} \geq [(x+1)(y+1)(z+1)]^{1 / 3}$

$\Rightarrow \frac{x+y+z+3}{3} \geq (x y z+x y+y z+x z+x+y+z+1)^{1 / 3}$

$\Rightarrow \quad\left(\frac{13}{3}ight)^3 \geq x y z+x y+y z+x z+11$

$\Rightarrow \quad \quad(y+1)(z+1)]^{1 / 3}$

Now, $x, y, z$ are integer.

$	herefore x y z+x y+y z+x z+11$ is also integer.

$	herefore\left(\frac{13}{3}ight)^3$ is also integer.

$	herefore \quad\left[\left(\frac{13}{3}ight)^3ight]=81\left[\because\left(\frac{13}{3}ight)^3=8137ight]$

$	herefore x y z+x y+y z+x z+11 \leq 81$

$\Rightarrow \quad x y z+x y+y z+x z \leq 70$

$	herefore$ Maximum value of $x y z+x y+y z+x z$ is

Let $x, y, z$ be three non-negative integers such that $x+y+z=10$. The maximum possible value of $x y z+x y+y z+z x$ is

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