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Question Number 144537 by loveineq last updated on 26/Jun/21

Let a, b > 0 and a + b = 2 . Prove that

(1) \frac{a^{3} + b^{3}}{2} - 2 (1 - a b) ⩾ 1

(2) \frac{a^{2} + b^{2}}{2} - 2 (1 - a b) ⩽ 1

Answered by mnjuly1970 last updated on 26/Jun/21

(1) : \frac{(a + b) (a^{2} + b^{2} - a b)}{2} - 2 + 2 a b

= a^{2} + b^{2} - 2 + a b = {(a + b)}^{2} - a b - 2

= 2 - a b \underset{i n e q u a l i t y}{\overset{a m - g m}{⩾}} 2 - \frac{a + b}{2} = 1 \dots

Commented by loveineq last updated on 26/Jun/21

t h a n k s

Answered by Olaf_Thorendsen last updated on 26/Jun/21

(1) : x = \frac{a^{3} + b^{3}}{2} - 2 (1 - a b)

x = \frac{{(a + b)}^{3} - 3 a b (a + b)}{2} - 2 (1 - a b)

x = \frac{8 - 6 a b}{2} - 2 (1 - a b)

x = 2 - a b = 2 - a (2 - a)

x = a^{2} - 2 a + 2 = {(a - 1)}^{2} + 1 ⩾ 1

(2) y = \frac{a^{2} + b^{2}}{2} - 2 (1 - a b)

y = \frac{(a + b) - 2 a b}{2} - 2 (1 - a b)

y = \frac{2 - 2 a b}{2} - 2 (1 - a b)

y = a b - 1 = a (2 - a) - 1

y = - a^{2} + 2 a - 1 = - {(a - 1)}^{2} ⩽ 0 ⩽ 1

Commented by loveineq last updated on 26/Jun/21

t h a n k s