IIT JEE , AIEEE Forum - Mathematics , Algebra

hpudipeddi

If real numbers x,y,z satisfy (x^2/y^2)+(y^2/z^2)+(z^2/x^2)=x/y+y/z+z/x, then prove that x=y=z

hsbhatt

I am not able to view the image. Could someone who can view it kindly type out the expression please?

rudra.panda

Me too cannot see the image.......................Please some body help.

feynmann

simple

let x/y,y,/z,z/x be a,b,c .( assume that they are positive ( yet to be proved ) )

so a^2 +b^2 +c^2 = a+b+c.........(1)

abc=1........(2)

From cauchy scharz inequality (a^2 +b^2+c^2)>=(1/3)* (a+b+c)^2.................(3)

from (1)&(3) we get a+b+c<=3 ......(4)

Again from A.M.>=G.M.

(a+b+c)>=3.............(5)

so from(4)&(5 ) a+b+c=3

so A.M.=G.M.

Which means a=b=c

i.e. x=y=z

hsbhatt

Or directly apply the inequality $a^2+b^2+c^2 \ge ab+bc+ca$ with equality occurring only when a = b = c

This is an application of the Cauchy Schwarz Inequality as follows

$(a^2+b^2+c^2) (b^2+c^2+a^2) \ge (ab+bc+ca)^2 \\ \\<br/>\Rightarrow (a^2+b^2+c^2)^2 \ge (ab+bc+ca)^2 \\ \\<br/>\Rightarrow a^2+b^2+c^2 \ge ab+bc+ca$

Or you just note that

$a^2+b^2+c^2 -ab -bc-ca = \frac{1}{2} [(a-b)^2+(b-c)^2+(c-a)^2] \ge 0$