Overdose on X Y Z!
Let be three non negative integers such that . What is the maximum possible value of ?
This problem is not original.
The answer is 69.
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1 solution
This is another KVPY question.
We usually tend to see multiplication(xyz) and eliminate thoughts of one of them being one.
If we consider x=y=z, we get x=y=z=10/3 = 3.333.
It is mentioned that x,y,z are integers. So,
x,y,z should be close to 3.33 and we take 3,3,4 and we get the answer = 69.
Another way is to differentiate the required statement partially and solving them by equating each of those to zero.
By doing this we get x=y and we conclude that it is 3,3,4.
OR we can use AM-GM here.
Take x+1,y+1,z+1 and apply the inequality.
( 3 1 3 ) 2 − 1 1 ≥ x y z + x y + y z + z x .
equality holds good when x = y = z but x + y + z = 10 and x, y, z are integers.
So maximum value occurs when any two of x, y, z are equal to 3 and third is equal to 4.
So, the answer is 69.