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Can anyone assist me in figuring out the way to tackle the question

Let $x$ be a real number such that $x > 0$. Prove that $x + (9/x) ≥ 6$.

I understand that its true because, for example, let $x = 1$ is greater than $6$... but how would i prove this?

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marked as duplicate by Asaf Karagila, Hagen von Eitzen, lab bhattacharjee, Stefan Hansen, Alexander Gruber Feb 15 '13 at 8:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You could use induction –  MITjanitor Feb 15 '13 at 6:46

2 Answers 2

up vote 1 down vote accepted

Here is a different solution to those mentioned in @gnometorule's comment above:

Consider $f(x) = x + \frac{9}{x}$ on $(0,\infty)$. Since $\lim_{x \downarrow 0} f(x) = \lim_{x \to \infty} f(x) = \infty$, and $f$ is continuous, we see that $f$ has a minimum on $(0,\infty)$.

Since $f$ is smooth, we can differentiate to get $f'(x) = 1-\frac{9}{x^2}$. Setting $f'(x) = 0$ gives the unique (in $(0,\infty)$) solution $x = 3$, which must be the minimum. Hence $f(x) \geq f(3)$ for all $x \in (0,\infty)$. Since $f(3) = 6$, we have $x + \frac{9}{x} \geq 6$.

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$$x + \frac{9}{x} - 6 = \frac{1}{x} (x^2 - 6x + 9) = \frac{1}{x} (x-3)^2$$

Therefore, if $x > 0$, then $x + \frac{9}{x} - 6 \geq 0$.

More generally, we can think of this as a variation on completing the square where we modify the middle term rather than the constant term. e.g.

$$ 4x + \frac{16}{x} = \frac{4}{x} (x^2 + 2^2) = \frac{4}{x} ((x-2)^2 + 4x) = \frac{4}{x} (x-2)^2 + 16 $$

so if $x > 0$, then $4x + \frac{16}{x} \geq 16$. (with equality only when $x=2$)

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