Question

Let $x,y,z$ be three non-negative integers such $x+y+z=10$. The maximum possible value $xyz+xy+yz+zx$ is:

• A. $52$
• B. $64$
• C. $69$
• D. $73$

Answer 1

The correct option is C $69$

We know that $x+y+z=10$
Therefore $(x+1)+(y+1)+(z+1)=13$
We have to maximize $xyz+xy+yz+zx$
Now, $xyz+xy+yz+zx=xyz+xy+yz+xz+x+y+z+1-11=(x+1)(y+1)(z+1)-11$
Using A.M.-G.M. on $(x+1),(y+1)$ and $(z+1)$ we get
$\frac{(x+1)+(y+1)+(z+1)}{3}\ge \sqrt[3]{3(x+1)(y+1)(z+1)}$
$\frac{13}{3}\ge \sqrt[3]{3(x+1)(y+1)(z+1)}$
$81\ge (x+1)(y+1)(z+1)$
$xyz+xy+yz+zx=(x+1)(y+1)(z+1)-11\le 81-11$
Therefore $xyz+xy+yz+zx\le 69$