7Using 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.
6 Answers
1057seeing 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.....
7
71Correct answer is 16.
@Sigma
(x+y+z)2< 270
So, How min( x+y+z ) = 16 ?
1Given:x2+y2+z2≥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
7@Vivek: fINDING OUT (X+Y+Z)2 <270, THE NEAREST SMALLER PERFECT SQUARE IS 162.
Since x,y,z are integers the minimum value is 16.
As x=4 y=5 z=7 to satisfy the minimum value of (x2+y2+z2)