Докажите неравенство а) (b-3)^2>b(b-6) б) b^2+10>_ 2(4b-3) в) (a-5)^2 > a(a-10) г) a^2+12>_ 4(2a-1)

Предыдущий
вопрос Следующий
вопрос

a)
(b-3)^2 > b(b-6)
Expanding the left side:
b^2 - 6b + 9 > b^2 - 6b
Rearranging the terms:
9 > 0
Since 9 is greater than 0, the inequality is always true for any value of b.

б)
b^2 + 10 ≥ 2(4b-3)
Expanding the right side:
b^2 + 10 ≥ 8b - 6
Rearranging the terms:
b^2 - 8b + 16 ≥ 0
(b-4)^2 ≥ 0
Since the square of any real number is always greater than or equal to 0, the inequality is always true for any value of b.

в)
(a-5)^2 > a(a-10)
Expanding the left side:
a^2 - 10a + 25 > a^2 - 10a
Rearranging the terms:
25 > 0
Since 25 is greater than 0, the inequality is always true for any value of a.

г)
a^2 + 12 ≥ 4(2a-1)
Expanding the right side:
a^2 + 12 ≥ 8a - 4
Rearranging the terms:
a^2 - 8a + 16 ≥ 0
(a-4)^2 ≥ 0
Since the square of any real number is always greater than or equal to 0, the inequality is always true for any value of a.