This is wrong. First of all, if the numbers are real, the tuple (-1,-1,1), which verifies the assertion $abc=1$ is undefined over reals. But even that, take (1,1,1), we have $3\sqrt{2} \geq 2*3$, which is not true.

I'll set another problem: Find the $max_k$ so that for all non-negative reals $a,b,c$ with $abc=1$ verifies: $$\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} +\sqrt{c+\frac{1}{c}} \geq k( \sqrt{a}+ \sqrt{b}+ \sqrt{c})$$

mathisreal wrote: The real numbers $a, b, c$ with $abc = 1$ Show that: $\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} + \sqrt{c + \frac{1}{c}}$ ≥ $2(\sqrt{a} + \sqrt{b} + \sqrt{c})$ Let $a,b,c>0,ab+bc+ac=1$ . Prove that$$\sqrt{a+\frac{1}{a}}+\sqrt {b+ \frac{1}{b}} +\sqrt {c+\frac{1}{c}} \ge 2 (\sqrt {a}+\sqrt{b} +\sqrt {c})$$(Russia)

whiwho wrote: I'll set another problem: Find the $max_k$ so that for all non-negative reals $a,b,c$ with $abc=1$ verifies: $$\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} +\sqrt{c+\frac{1}{c}} \geq k( \sqrt{a}+ \sqrt{b}+ \sqrt{c})$$ Let $a=b=t,c=\frac{1}{t^2},t\to 0^+$ get $k_{max}=1$

Meaningless Sorry.

sqing wrote: Grotex wrote: whiwho wrote: I'll set another problem: Find the $max_k$ so that for all non-negative reals $a,b,c$ with $abc=1$ verifies: $$\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} +\sqrt{c+\frac{1}{c}} \geq k( \sqrt{a}+ \sqrt{b}+ \sqrt{c})$$ Let $a=b=t,c=\frac{1}{t^2},t\to 0^+$ get $k_{max}=1$ For all positive reals $a,b,c$ with $abc=1$ , prove that $$\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} +\sqrt{c+\frac{1}{c}} >\sqrt{a}+ \sqrt{b}+ \sqrt{c}$$ Obvious($\sqrt{a + \frac{1}{a}}>\sqrt{a}$)

What the hack wrong with this problem $\frac{1}{a}>0$.......and we re done.......

The real problem was: The positive real numbers $a, b, c$ with $ab+bc+ca = 1$ Show that: $\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} + \sqrt{c + \frac{1}{c}}$ ≥ $2(\sqrt{a} + \sqrt{b} + \sqrt{c})$

Mathsy123 wrote: The real problem was: The positive real numbers $a, b, c$ with $ab+bc+ca = 1$ Show that: $\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} + \sqrt{c + \frac{1}{c}}$ ≥ $2(\sqrt{a} + \sqrt{b} + \sqrt{c})$ Solution. Note that \[ \sum_{cyc} \sqrt{a + \frac{1}{a}} = \sum_{cyc} \sqrt{ \frac{a^2 + 1}{a}}= \sum_{cyc} \sqrt{\frac{(a + b)(a + c)}{a}} \ge \sum_{cyc} \sqrt{a} + \sqrt{\frac{bc}{a}} \]It's sufficient to prove that \[ \sum_{cyc} \sqrt{\frac{bc}{a}} \ge \sum_{cyc} \sqrt{a} \]Which is, immediately true by AM-GM. \[ \sum_{cyc} bc \ge \sum_{cyc} a\sqrt{bc} \]