Numerele pozitive \( x_{1}, x_{2}, \ldots , x_{n} \) satisfac identitatea
\( \frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\ldots +\frac{1}{x_{n}+1}=1. \)
Sa se demonstreze ca
\( \sqrt{x_{1}}+\sqrt{x_{2}}+\ldots \sqrt{x_{n}}\geq(n-1)\left(\frac{1}{\sqrt{x_{1}}}+\frac{1}{\sqrt{x_{2}}}+\ldots +\frac{1}{\sqrt{x_{n}}}\right) \).
"Vojtech Jarnik", 2002, Categoria I
Inegalitate in n variabile cu radicali
Moderators: Laurian Filip, Filip Chindea, Radu Titiu, maky, Cosmin Pohoata
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Inegalitate in n variabile 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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La inceput poate am fi condusi spre idei mai elevate (poate si de sursa problemei), dar o scurta analiza spune altceva. Mai intai, evident ca acel \( n - 1 \) face inegalitatea nenaturala; adunând \( \sum \frac{1}{\sqrt{x_k}} \) in ambii membrii, avem de aratat \( n \cdot \sum \frac{1}{\sqrt{x_k}} \le \sum \left( \sqrt{x_k} + \frac{1}{\sqrt{x_k}} \right) = \sum \frac{x_k + 1}{\sqrt{x_k}} \). Ideea de "omogenizare" este instantanee, scriind membrul drept ca \( \left( \sum \frac{1}{x_k + 1} \right) \left( \sum \frac{x_k + 1}{\sqrt{x_k}} \right) \). Ramanem cu
\( n \cdot \sum \left( \frac{1}{x_k + 1} \cdot \frac{x_k + 1}{\sqrt{x_k}} \right) \le \left( \sum \frac{1}{x_k + 1} \right) \left( \sum \frac{x_k + 1}{\sqrt{x_k}} \right) \),
si s-a terminat, deoarece \( a, b > 0 \), \( ab \ge 1 \) implica
\( \left( \frac{1}{a + 1} - \frac{1}{b + 1} \right) \left( \frac{a + 1}{\sqrt{a}} - \frac{b + 1}{\sqrt{b}} \right) = \frac{(1 - \sqrt{ab})(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})^2}{\sqrt{ab}(a + 1)(b + 1)} \le 0 \),
si cum pentru orice \( x_k \neq x_m \), \( \frac{1}{x_k + 1} + \frac{1}{x_m + 1} \le 1 \Rightarrow x_kx_m \ge 1 \), totul rezulta din inegalitatea lui Cebasev pentru \( n \)-uple invers ordonate.
\( n \cdot \sum \left( \frac{1}{x_k + 1} \cdot \frac{x_k + 1}{\sqrt{x_k}} \right) \le \left( \sum \frac{1}{x_k + 1} \right) \left( \sum \frac{x_k + 1}{\sqrt{x_k}} \right) \),
si s-a terminat, deoarece \( a, b > 0 \), \( ab \ge 1 \) implica
\( \left( \frac{1}{a + 1} - \frac{1}{b + 1} \right) \left( \frac{a + 1}{\sqrt{a}} - \frac{b + 1}{\sqrt{b}} \right) = \frac{(1 - \sqrt{ab})(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})^2}{\sqrt{ab}(a + 1)(b + 1)} \le 0 \),
si cum pentru orice \( x_k \neq x_m \), \( \frac{1}{x_k + 1} + \frac{1}{x_m + 1} \le 1 \Rightarrow x_kx_m \ge 1 \), totul rezulta din inegalitatea lui Cebasev pentru \( n \)-uple invers ordonate.
Life is complex: it has real and imaginary components.