If x,y,z are any three positive numbers, then show that

x^2/y + y^2/z + z^2/x >= x+y+z.

Hi Speed Racer,

Consider,

x²/y + y, and apply AM-GM inequality:

You will get

x²/y + y ≥ 2x

Apply this cyclically, and add, you will get (x²/y+y)+(y²/z+z)+(z²/x+x)≥2(x+y+z)

And hence the given inequality: (x²/y)+(y²/z)+(z²/x)≥(x+y+z).

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).

LHS=X²/y+y²/z+z²/x

     =x³z+y³x+z³y/ xyz     ( suppose put min. value of x,y,z i.e= 0)

    then it is ∞ and when putiing any other value of those i.e. 1 then it becomes 3 i.e  equal to RHS.

                                or

   put the AM- GM inequality then it also clear this statement.!!!

       if u like it plz approve it!!!!