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How to find the smallest value of $a$ that a solution for the following inequality exists?

$$2\ln x + \frac{5}{x} < a$$

These are the steps I have done, and I have no idea what to do next:

\begin{align*} 2\ln x &< -\frac{5}{x} + a\\ \ln x^2 &< -\frac{5}{x} + a\\ e^{-\frac{5}{x} + a} &< x^2\\ e^{-5} \cdot e^{\frac{1}{x}} \cdot e^a &< x^2 \end{align*}

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  • $\begingroup$ $u<v$ if and only if $e^u<e^v$. You've instead concluded $e^v<e^u$. $\endgroup$
    – Thomas Andrews
    Jan 23, 2013 at 12:52

1 Answer 1

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As Thomas Andrews has pointed out, the inequalities in your third and fourth lines are the wrong way around. Having said that, I'm not sure whether your approach will work even if you make these corrections. Instead I present an alternative method which uses a little bit of differential calculus.

Let $f(x) = 2\ln x + \frac{5}{x}$ and note that it has domain $(0, \infty)$. The only way the inequality $f(x) < a$ can't have solutions is if $f(x) \geq a$ for all $x \in (0, \infty)$. In particular, if we know that $f$ has a global minimum (it need not have one), then for any $a$ less than or equal to that, the inequality will have no solutions.

You should be able to use calculus to determine whether $f$ has a global minimum, and if so, what it is.

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