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Let $k$ be a real number such that the inequality, $\sqrt{x-1}+\sqrt{4-x}>k$ has a solution. What is the maximum value of $k$.

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2 Answers

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$$\sqrt{x-1}+\sqrt{4-x}=\sqrt{3+2\sqrt{(x-1)(4-x)}}\geq\sqrt3.$$ The equality occurs for $x=1$.

Thus, $\sqrt3$ is a minimal value of $\sqrt{x-1}+\sqrt{4-x}.$

The maximal value we can get by C-S: $$\sqrt{x-1}+\sqrt{4-x}\leq\sqrt{(1^2+1^2)(x-1+4-x)}=\sqrt{6}.$$ The equality occurs for $x=2.5.$

Thus, all $k<\sqrt{6}$ is valid and the maximal value of $k$ does not exist.

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Hint: The expression $\sqrt{x-1}+\sqrt{4-x}$ only makes sense when $x\in[1,4]$. What's the graph of the map$$\begin{array}{ccc}[1,4]&\longrightarrow&\mathbb R\\x&\mapsto&\sqrt{x-1}+\sqrt{4-x}?\end{array}$$

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