IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

Certain elements may kindly keep off (you know who I mean):

$ext{find real numbers} x_1, x_2, ..., x_n ext{satisfying} \ \ sqrt{x_1 - 1^2} + 2sqrt{x_2 - 2^2} + 3sqrt{x_3 - 3^3} + ... + n sqrt{x_n - n^2} = rac{1} {2} (x_1 + x_2+...+x_n)$

typo: that is 3² not 3³

nadeemoidu

We are waiting for you to open the question for all elements

sandeepramesh

yes that's unfair to 'those certain elements' :D Well if u tell what elements we can ask the other elements of the body to post the answer :P

anchitsaini

well i am not getting the elegant way out but nevertheless --

1)taking n=1

(x₁ - 1²) = x1/2

this gives x₁ = 2 = 2 * 1²

2) taking n=2

we get similarly x₂ = 8 = 2 * 2²

3) taking n=3

we get x₃= 18= 2 * 3²

now we seem to get a series --

1² + 2² + 3² +.....

thus

(x _n - n² )= n

hence the next terms would be

x_n = 2n²

hsbhatt

ok now its open. and i sincerely thank "certain elements" for their kind cooperation.

the answer anchit has got is right. now for the other approaches.

nadeemoidu

By AM-GM inequality , (k² + x_k - k² )/2 >= k

(x_k - k²)

=> x_k/ 2 >= k

(x_k - k²)

Sum this for all values of k we get the given expression , with the equality sign.
This means that equality holds for each equation.

Equality holds for AM-GM inequality when , the 2 numbers are equal

So k² = x_k - k²

x_k =2 k²

hsbhatt

nice (as expected from nadeem).

sandeepramesh

another method that struck me immediately at that time was this
EDIT: There was a typo, it shd be sec^2yi

Put $x_{i} = i^2 cdot sec {y_{i}}$

$implies 1^2 cdot an {y_{1}} + 2^2 cdot an {y_{2}} + cdots + n^2 cdot an {y_{n}} =$
$rac {1}{2} cdot (1^2 cdot sec^2 {y_{1}} + 2^2 cdot sec^2 {y_{2}} +cdots+ n^2 cdot sec^2 {y_{n}})$

$implies 1^2 cdot an {y_{1}} + 2^2 cdot an {y_{2}} + cdots + n^2 cdot an {y_{n}} =$
$rac {1}{2} cdot ( 1^2 cdot (1 + an ^2 {y_{1}}) + 2^2 cdot (1 + an ^2 {y_{2}}) + cdots + n^2 cdot (1 + an ^2 {y_{n}}))$

as tanyi can be random, compare coeffs to get tanyi = 1 and hence xi = i^2 :)