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If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$.

Working:

$\frac{1}{2}(1-a)\leq 0<\frac{1}{2}(b-a)$ and $(1-\sqrt{a})\leq 0<\sqrt{b}-\sqrt{a}$

Trying to show $\frac{1}{2}(b-a)-(\sqrt{b}-\sqrt{a})$ is positive but don't know what to do next.

shredalert
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3 Answers3

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Note that $$ b-a = (\sqrt b - \sqrt a)(\sqrt b + \sqrt a) $$ and $\sqrt b + \sqrt a >1+1$.

Arthur
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  • Solution: $\frac{1}{2}(b-a)(\sqrt{b}-\sqrt{a})=(\frac{1}{2}(\sqrt{b}+\sqrt{a})-1)(\sqrt{b}-\sqrt{a})$. $\frac{1}{2}\leq\frac{1}{2}\sqrt{b}$ and $\frac{1}{2}\leq\frac{1}{2}\sqrt{a}$ $\Rightarrow 1<\frac{1}{2}\sqrt{b}+\frac{1}{2}\sqrt{a}$ according to your very helpful hint. Thus $\frac{1}{2}(b-a)(\sqrt{b}-\sqrt{a})=(\frac{1}{2}(\sqrt{b}+\sqrt{a})-1)(\sqrt{b}-\sqrt{a})$ is positive. – shredalert Nov 29 '16 at 22:10
  • One more identity added to the arsenal. This makes me happy. :D – shredalert Nov 29 '16 at 22:22
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One may observe that, for $b>a\ge1$, $$ \begin{align} (b-a)-2(\sqrt{b}-\sqrt{a})&=(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})-2(\sqrt{b}-\sqrt{a}) \\\\&=(\sqrt{b}-\sqrt{a})\left[\sqrt{b}+\sqrt{a}-2 \right] \\\\&=(\sqrt{b}-\sqrt{a})\left[(\sqrt{b}-1)+(\sqrt{a}-1) \right] \\\\&>0 \end{align} $$ by using that $x \mapsto \sqrt{x}$ is strictly increasing over $[1,\infty)$.

Olivier Oloa
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0

hint: Use MVT for $f(x) = \sqrt{x} $

DeepSea
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