An exercise from the Mathematics Olympiad with a solution , Prove that x²/y+y²/z+z²/x ≥ x+y+z

 

x and y and z are positive real numbers

Prove that :

solution:

we have (x-y)² ≥ 0

Means x²-2xy+y² ≥ 0 , Means x²+y² ≥ 2xy  ,

 Means  (x²+y²)/y ≥ 2xy/y , then  x²/y+y  ≥ 2x   (1)

In the same way we show that :  y²/z+z  ≥ 2y   (2)

                                                     z²/x+x ≥ 2z   (3)

We collect the inequalities 1 and 2 and 3 

 x²/y+y + y²/z+z + z²/x+x ≥ 2x+2y+2z

   so:

               x²/y + y²/z + z²/x ≥ x+y+z

    

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