Prove that a^2-4a+b^2+10b+29>=0 if a,b are real numbers .

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We have to prove that a^2 - 4a + b^2 + 10b + 29>=0, for real values of a and b.

a^2 - 4a + b^2 + 10b + 29

=> a^2 - 4a + 4 + b^2 + 10b + 25

=> (a - 2)^2 + (b + 5)^2

The sum of squares of real numbers is always positive.

This proves that a^2 - 4a + b^2 + 10b + 29 >= 0

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