Question

Determine all natural numbers x, y and z such that x ≤ y ≤ z, and xyz + xy + xz + yz + x + y + z + 1 = 2020

Answer 1

Answer:

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

3

(x+1)+(y+1)+(z+1)

≥

3

(x+1)(y+1)(z+1)

3

13

≥

3

(x+1)(y+1)(z+1)

81≥(x+1)(y+1)(z+1)

xyz+xy+yz+zx=(x+1)(y+1)(z+1)−11≤81−11

Therefore xyz+xy+yz+zx≤69

Step-by-step explanation:

Hope it is helpful for you.