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Question Number 136739 by Ar Brandon last updated on 25/Mar/21

Given 0 < a < b, prove that

\frac{{(b - a)}^{2}}{8 b} ⩽ \frac{a + b}{2} - \sqrt{ab} ⩽ \frac{{(b - a)}^{2}}{8 a}

Answered by snipers237 last updated on 26/Mar/21

\frac{a + b}{2} - \sqrt{a b} = \frac{{(\sqrt{a} - \sqrt{b})}^{2}}{2} a n d b - a = (\sqrt{b} - \sqrt{a}) (\sqrt{b} + \sqrt{a})

0 < a < b \Rightarrow \sqrt{a} < \sqrt{b} . T h e n 2 \sqrt{a} < \sqrt{a} + \sqrt{b} < 2 \sqrt{b}

S o 4 a < {(\sqrt{a} + \sqrt{b})}^{2} < 4 b

\frac{{(b - a)}^{2}}{8 b} = \frac{{(\sqrt{b} - \sqrt{a})}^{2}}{2} . \frac{{(\sqrt{b} + \sqrt{a})}^{2}}{4 b} < \frac{{(\sqrt{a} - \sqrt{b})}^{2}}{2}

\frac{{(b - a)}^{2}}{8 a} = \frac{{(\sqrt{b} - \sqrt{a})}^{2}}{2} . \frac{{(\sqrt{b} + \sqrt{a})}^{2}}{4 a} > \frac{{(\sqrt{a} - \sqrt{b})}^{2}}{2}

L F Y T C

Commented by Ar Brandon last updated on 26/Mar/21

Thanks