Question

X^2+y^2-2(y+2x)+5≥0​

Answer 1

x^2+y^2-2(y+2x)+5≥0​

x^2+y^2-2y-4x+1+4≥0​

x^2-4x+4+y^2-2y+1≥0​

(x-2)^2 + (y-1)^2 ≥0​

Сумма двух квадратов всегда неотрицательна

выполняется при любых x y

= 0 при x = 2 y = 1

Answer 2

The given inequality is:

x^2 + y^2 - 2(y + 2x) + 5 ≥ 0

To analyze this inequality, let's rearrange the terms:

x^2 + y^2 - 2y - 4x + 5 ≥ 0

Now, let's try to simplify it further. To do that, we'll group the x and y terms separately:

(x^2 - 4x) + (y^2 - 2y) + 5 ≥ 0

Now, complete the square for both x and y terms separately:

(x^2 - 4x + 4) + (y^2 - 2y + 1) + 5 - 4 - 1 ≥ 0

(x - 2)^2 + (y - 1)^2 ≥ -1

Now, since (x - 2)^2 and (y - 1)^2 are both squares of real numbers, they are always non-negative. Therefore, the sum of two non-negative terms cannot be less than -1. In other words, the expression (x - 2)^2 + (y - 1)^2 ≥ -1 is always true for any values of x and y.

Hence, the given inequality x^2 + y^2 - 2(y + 2x) + 5 ≥ 0 holds true for all values of x and y.