MHB Need Explanations For The Red Part (2 variable inequality)

AI Thread Summary
The discussion focuses on proving the inequality [(x^2 + y^2)/2] >= x + y - 1 for any two numbers x and y. The key argument is based on the fact that the square of any real number is non-negative, leading to the conclusion that (x-1)^2 + (y-1)^2 is also non-negative. This non-negativity implies that the left side of the inequality is greater than or equal to the right side. The participants seek clarification on the reasoning behind the red part of the proof, which hinges on understanding the properties of squares. Overall, the discussion emphasizes the foundational principles of inequalities in algebra.
WannaBe
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Prove that for any two numbers x,y we have [(x^2 + y^2)/2] >= x + y - 1

Solution)

For any number a we have have a^2 > 0. So,

(x-1)^2 + (y-1)^2 >= 0

And if we solve this we get the solution.I don't get the red part.
 
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Re: Need Explanations For The Red Part

WannaBe said:
Prove that for any two numbers x,y we have [(x^2 + y^2)/2] >= x + y - 1
Solution)
For any number a we have have a^2 > 0. So,
(x-1)^2 + (y-1)^2 >= 0

And if we solve this we get the solution.
If for all $$a$$ we have $$a^2\ge 0$$ then $$(x-1)^2\ge 0~\&~(y-1)^2\ge 0$$.

Add those to together and get $$(x-1)^2+(y-1)^2\ge 0$$
 
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