Question

1) Докажите неравенства (b-1)(b-3)<(b-2)^22) Докажите a^2+b^2-6a+2b+10 больше или равно нульпдоробно пожалуйста)

Answer 1

(b-1)(b-3)<(b-2)^2   b^2-4b+3<b^2-4b+4  3<4

 

a^2+b^2-6a+2b+10=a^2-6a+9+b^2+ 2b+1=(a-3)^2+(b+1)^2≥0

Answer 2

Proof of Inequality (b-1)(b-3) < (b-2)^2

To prove the inequality (b-1)(b-3) < (b-2)^2, let's start by expanding and simplifying the left-hand side of the inequality.

Expanding (b-1)(b-3): (b-1)(b-3) = b^2 - 3b - b + 3 = b^2 - 4b + 3

Now, let's simplify the right-hand side of the inequality.

Expanding (b-2)^2: (b-2)^2 = (b-2)(b-2) = b^2 - 2b - 2b + 4 = b^2 - 4b + 4

Comparing the two expressions, we have: (b-1)(b-3) = b^2 - 4b + 3 (b-2)^2 = b^2 - 4b + 4

We can see that (b-1)(b-3) is less than (b-2)^2 by 1, as the constant term in (b-2)^2 is greater than the constant term in (b-1)(b-3).

Therefore, we have proved that (b-1)(b-3) < (b-2)^2.

Proof of Inequality a^2 + b^2 - 6a + 2b + 10 ≥ 0

To prove the inequality a^2 + b^2 - 6a + 2b + 10 ≥ 0, let's complete the square for the terms involving 'a' and 'b'.

For the 'a' terms: a^2 - 6a = (a^2 - 6a + 9) - 9 = (a - 3)^2 - 9

For the 'b' terms: b^2 + 2b = (b^2 + 2b + 1) - 1 = (b + 1)^2 - 1

Now, substituting these expressions back into the original inequality: (a^2 - 6a + 9) + (b^2 + 2b + 1) - 9 - 1 + 10 ≥ 0 ((a - 3)^2 - 9) + ((b + 1)^2 - 1) ≥ 0 (a - 3)^2 + (b + 1)^2 ≥ 0

The square of any real number is non-negative, and the sum of two non-negative numbers is also non-negative. Therefore, the inequality a^2 + b^2 - 6a + 2b + 10 ≥ 0 holds true for all real values of 'a' and 'b'.

Hence, we have proved that a^2 + b^2 - 6a + 2b + 10 ≥ 0.