Minimum of x+y+z

seeing the condition u posted.....to keep x+y+z min i shud keep x and y as small as possible....
keeping it 4 and 5 i get z=7
so x+y+z=16
i know my solution is not up to the mark....but good for MCQ exams like JEE.....

Using C.S.inequality:

(ax+by+cz)²< equal to (a²+b²+c²) (x²+y²+z²).
Here a=b=c=1.

so (x+y+z)²<equal to 3 x 90.(taking the min value)

<equal to 270

Since x, y , z are integers nearest perfect square is 256.

So, x+y+z=16.

Correct answer is 16.

@Sigma

(x+y+z)²< 270

So, How min( x+y+z ) = 16 ?

Given:x²+y²+z²â‰¥90----------------------(1)
→ min of this expression is 90
also given that x≥4
y≥5
z≥6
equation (1) gets min value at x=4 , y=5 , z=7
so min of x+y+z=4+5+7=16.
Ans: 16

It is 16
using derivative