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For $0<b \leq a$ prove that $$\frac{1}{8}\left(\frac{(a-b)^2}{a} \right) \leq \frac{a+b}{2}-\sqrt{ab} \leq \frac{1}{8}\left(\frac{(a-b)^2}{b} \right) $$

I was trying compare the following expressions $$\frac{a+b}{2}-\sqrt{ab} =\frac{1}{2}(\sqrt{a}-\sqrt{b})^2$$

$$\frac{1}{8}\left(\frac{(a-b)^2}{a} \right)=\frac{1}{2a}\left(\frac{a-b}{2}\right)^2\leq\frac{1}{a}(a-b)^2$$ and use the fact that $\sqrt{a} \leq a$ when $a>0$ but I don´t get good results.

Any hint of how I should start?

Juan T
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  • Hint:

    i) $a+b - 2\sqrt{ab} = (\sqrt a - \sqrt b)^2$

    ii) $a-b = (\sqrt a - \sqrt b) (\sqrt a + \sqrt b)$

    iii) $2\sqrt b \leq \sqrt a + \sqrt b \leq 2\sqrt a$

     – Math Lover Mar 06 '21 at 19:27
  • Why did you add the tag multivariable-calculus? – Yagger Mar 12 '21 at 20:37
  • Becasue usually problems of show inequalitys have nice solution using tools of multivariable calculus – Juan T Mar 12 '21 at 20:56

1 Answers1

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Hint: $$ \frac{1}{2}(\sqrt a - \sqrt b )^2 = \frac{b}{{2(\sqrt a + \sqrt b )^2 }}\left( {\frac{{(a - b)^2 }}{b}} \right) $$ and $$ \frac{1}{2}(\sqrt a - \sqrt b )^2 = \frac{a}{{2(\sqrt a + \sqrt b )^2 }}\left( {\frac{{(a - b)^2 }}{a}} \right). $$ Use the fact that $0<b\leq a$ to estimate the right-hand sides from above and below.

Gary
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  • I think that it could useful, how do you think in this equalty? – Juan T Mar 06 '21 at 19:18
  • @JuanT Are you asking about the way I got these equalities? – Gary Mar 06 '21 at 19:25
  • Yes, about the creative process for get it – Juan T Mar 06 '21 at 19:26
  • $$ \frac{1}{2}(\sqrt a - \sqrt b )^2 = \frac{1}{2}\frac{{(\sqrt a - \sqrt b )^2 (\sqrt a + \sqrt b )^2 }}{{(\sqrt a + \sqrt b )^2 }} = \frac{1}{2}\frac{{\left[ {(\sqrt a - \sqrt b )(\sqrt a + \sqrt b )} \right]^2 }}{{(\sqrt a + \sqrt b )^2 }} = \frac{1}{2}\frac{{(a - b)^2 }}{{(\sqrt a + \sqrt b )^2 }} = \cdots $$ – Gary Mar 06 '21 at 19:26
  • great, thanks a lot, nice hint. – Juan T Mar 06 '21 at 19:29