LIX OM - I - Task 9

Determine the smallest real number a with the following property: 
For any real numbers $ x, y, z \geqslant a $ satisfying the condition $ x + y + z = 3 $
the inequality holds 
Answer: $ a = -5 $. 
In the solution, we will use the following identity: 






Suppose that the number $ a \leqslant 1 $ has the property given in the problem statement. The numbers
$ x = a, y = z = 2\cdot (\frac{3-a}{2}) $ satisfy the conditions $ x, y, z \geqslant a $ and $ x+y+z = 3 $, thus






by virtue of (1), from which the inequality $ a \geqslant -5 $ follows. 
We will now show that the number $ a = -5 $ satisfies the conditions of the problem. For this purpose, consider numbers
$ x, y, z \geqslant-5 $ such that $ x + y + z = 3 $. Without loss of generality, assume that
$ x \leqslant y \leqslant z $. The sum $ y + z $ is a positive number, because otherwise, given
$ y\leqslant z $, we would have $ x \leqslant y \leqslant 0 $, which would imply that $ x +y + z \leqslant y + z \leqslant 0 $,
contradicting the condition $ x+y+z = 3 $. Since $ y + z > 0 $, the inequality






holds, and from the identity (1) we finally obtain 














(Note: The last part of the text seems to be incomplete or missing in the original Polish text, so the translation reflects that.)