$$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ I know it is trivial to prove straight forward but I need to prove it using AM-GM inequality.
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Why do you assume it can be done? – Thomas Andrews Nov 14 '14 at 23:31
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I don't know it for sure. This exercise in a math book an i assume that it can be done. – Bakyr Nov 14 '14 at 23:34
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1Please state the exercise exactly. – Thomas Andrews Nov 14 '14 at 23:35
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1It's already stated. Prove the following ... using AM-GM inequality. That's it. – Bakyr Nov 14 '14 at 23:39
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@Bakyr Even adding the harmonic mean inequality doesn't seem to help in this case. Were you specifically asked to prove this with the AM-GM inquality or is it only a personal quest? – Timbuc Nov 14 '14 at 23:44
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1You can delete comments, @Mick – Thomas Andrews Nov 15 '14 at 02:08
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@ThomasAndrews How can I do that (after 4 minutes)? – Mick Nov 16 '14 at 04:43
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The delete comments "x" mark, when you hover over your comment, always works. @Mick – Thomas Andrews Nov 16 '14 at 05:51
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\begin{align*} \sqrt a + \sqrt b &= \frac{(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}) + (\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab})}{2} \\ &\ge \sqrt{\big(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}\big) \big(\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab}\big)} \\ &= \sqrt{\big(\sqrt a + \sqrt b\big)^2 - 2 \sqrt{ab}} \\ &= \sqrt{a+b} \end{align*} (To apply AM/GM thus requires $\sqrt a + \sqrt b \ge \sqrt2\sqrt[4]{ab}$; happily, $\sqrt a + \sqrt b \ge 2\sqrt[4]{ab}$, by AM/GM again.)