Fie \( a, b, c \) trei numere reale strict pozitive astfel incat \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\sqrt{abc} \). Sa se arate ca \( abc\geq\sqrt{3(a+b+c)} \).
Cezar Lupu, R.M.I. C-ta, 2004
Inegalitatea 4, cu radicali
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Cezar Lupu
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Inegalitatea 4, cu radicali
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Filip Chindea
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One-line solution, folosind binecunoscuta \( ab+bc+ca \ge \sqrt{3abc(a+b+c)} \):
\( \sqrt{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + bc + ca}{abc} \Rightarrow abc = \frac{ab + bc + ca}{\sqrt{abc}} \ge \frac{\sqrt{3abc(a+b+c)}}{\sqrt{abc}} = \sqrt{3(a+b+c)} \).
\( \sqrt{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + bc + ca}{abc} \Rightarrow abc = \frac{ab + bc + ca}{\sqrt{abc}} \ge \frac{\sqrt{3abc(a+b+c)}}{\sqrt{abc}} = \sqrt{3(a+b+c)} \).
Life is complex: it has real and imaginary components.