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- Thread starter WannaBe
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If for all \(\displaystyle a\) we have \(\displaystyle a^2\ge 0\) then \(\displaystyle (x-1)^2\ge 0~\&~(y-1)^2\ge 0\).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.

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