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Question Number 193138 by Mastermind last updated on 04/Jun/23

Show that for all a, b \in R

i) ab ⩽ \frac{1}{2} (a^{2} + b^{2})

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

iii) \sqrt{ab} ⩽ \frac{1}{2} (a + b), for a, b ⩾ 0 such that

a and b have square roots .

Help

Answered by Subhi last updated on 04/Jun/23

(i) {(a - b)}^{2} ⩾ 0

a^{2} + b^{2} - 2 a b ⩾ 0

a b ⩽ \frac{1}{2} (a^{2} + b^{2})

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

n o t e t h a t 2 a b ⩽ a^{2} + b^{2}

a^{2} + b^{2} + 2 a b ⩽ 2 (a^{2} + b^{2})

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

(i i i) {(\sqrt{a} - \sqrt{b})}^{2} ⩾ 0

a + b - 2 \sqrt{a b} ⩾ 0

\sqrt{a b} ⩽ \frac{1}{2} (a + b)

Answered by aba last updated on 04/Jun/23

i) {(a - b)}^{2} ⩾ 0 \Rightarrow \frac{a^{2} + b^{2}}{2} ⩾ ab

ii) ab = \frac{1}{2} ({(a + b)}^{2} - (a^{2} - b^{2})) ⩽ \frac{a^{2} + b^{2}}{2} \Rightarrow \frac{(a^{2} + b^{2})}{2} ⩽ a^{2} + b^{2}

iii) {(\sqrt{a} - \sqrt{b})}^{2} ⩾ 0 \Rightarrow ab ⩽ \frac{a + b}{2}

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