*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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