a^2 + b^2 + 4a - 2b+5 >= 0

Let us rewrite:

==> a^2 +4a + b^2 - 2b >= -5

Let us add 4 to both sides:

==> a^2 + 4a + 4 + b^2 - 2b >= -1

==> (a+2)^2 + b^2 - 2b >= -1

Now add 1 to both sides:

==> (a+ 2)^2 + b^2 - 2b + 1 >= 0

==> (a+2)^2 + (b-1)^2 >= 0

But (a+2)^2 >=0

and (b-1)^2>= 0

Then The inequality is true.

To prove a^2+b^2 +4a-2b+5 > = 0

Solution:

Rearrange the LHS as:

(a^2+4a+4 ) + (b^2-2b+1)

(a+2)^2+(b-1)^2 is the sum of two squares which is always positive and could be zero only when a = -2 and b = 1

So a^2+b^2 +4a-2b+5 > = 0

We'll re-write the terms so that to complete the squares:

a^2 + b^2 + 4a - 2b + 5 = a^2 + b^2 + 4a - 2b + 4 + 1

We notice that we've substituted the value 5 by the sum 4+1.

We'll collect the terms to complete the square:

a^2+4a+4 = (a+2)^2

We'll collect the rest of the terms to complete another square:

b^2 - 2b + 1 = (b-1)^2

We'll re-write the inequality:

(a+2)^2 + (b-1)^2 >=0

Since the sum of 2 square is positive, the inequality is verified.

The sum of squares will be zero if and only if:

a+2 = 0

We'll subtract 2 both sides:

**a = -2**

and

b-1 = 0

We'll subtract 1 both sides:

**b = 1**