Need Explanations For The Red Part (2 variable inequality)

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SUMMARY

The discussion focuses on proving the inequality \(\frac{x^2 + y^2}{2} \geq x + y - 1\) for any two numbers \(x\) and \(y\). The proof utilizes the fact that for any number \(a\), \(a^2 \geq 0\), leading to the conclusion that \((x-1)^2 + (y-1)^2 \geq 0\). This establishes the validity of the inequality by demonstrating that the sum of squares is always non-negative.

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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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