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

Alll we need to do is to express the left sides as a sum
of postivie numbers.

We'll create perfect squares to the
left side:

(a^2 - 4a + 4) + (b^2 + 10b + 25) - 4 - 25 + 29
>=0

We notice that we've added the numbers 4 and 25
to complete the squares. For the inequality to hold, we'll have to subtract these added
values.

(a - 2)^2 + (b + 5)^2 - 29 + 29 >=
0

We'll eliminate like
terms:

(a - 2)^2 + (b + 5)^2 >
=0

Since the squares are always positive, the
inequality is verified, for any real values of a and
b.