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Suppose $x>y>5$. Prove that: $$\sqrt{x}-\sqrt{y} \leqslant \frac{1}{4}(x-y).$$

This is an exercise in a textbook I am currently working through in preparation for college (it will be used in a course). I am just looking for hints and pointers on how to start. Apologies for the poor formatting.

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Then $$\begin{align*} x - y &= (\sqrt x + \sqrt y)(\sqrt x - \sqrt y)\\ &> 2\sqrt 5(\sqrt x - \sqrt y)\\ &> 4(\sqrt x - \sqrt y)\\ \sqrt x - \sqrt y &<\frac14(x-y) \end{align*}$$

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We need to prove that $\sqrt{x}+\sqrt{y}\geq4$, which is true because $$\sqrt{x}+\sqrt{y}>\sqrt5+\sqrt5>4.$$

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Observe that $$\frac{\sqrt x-\sqrt y}{x-y}=\frac1{\sqrt{x}+\sqrt{y}}$$

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$$x>y>5 \implies \sqrt{x}>\sqrt{y}>\sqrt{5}$$ $$\implies \sqrt{x}-\sqrt{y}=\frac{x-y}{\sqrt{x}+\sqrt{y}}< \frac{x-y}{2\sqrt{5}}<\frac{x-y}{4}$$

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I think the sign must be reverse.

Hints $$a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}).$$ $$\sqrt{x}>\sqrt{y}>2.$$

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The derivative of $f(u) = \sqrt u$ is $$f'(u) = \frac{1}{2\sqrt u}$$

If $u > 5$, then $f'(u) < \frac{1}{2\sqrt5} < \frac14$.

By mean value theorem, for $x > y > 5$, there exist some $c$ such that $x > c > y > 5$ and

$$\begin{align*} \frac{\sqrt x - \sqrt y}{x - y} &= f'(c)\\ \frac{\sqrt x - \sqrt y}{x - y} &< \frac 14\\ \sqrt x - \sqrt y &< \frac14(x - y) \end{align*} $$

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