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Sa se demonstreze inegalitatea:
\( \frac{C^0_n}{1}+\frac{C^1_n}{3}+\frac{C^2_n}{5}+\dots+\frac{C^n_n}{2n+1} \geq \frac{2^{n+1}}{2n+1}, \) \( n \in N^* \).

Virgil Nicula,GM 7-8/1999

Notam \( S_n=\displaystyle\frac{C^0_n}{1}+\frac{C^1_n}{3}+\frac{C^2_n}{5}+\dots+\frac{C^n_n}{2n+1}\geq\frac{2^{n+1}}{2n+1} \)
Cautam o relatie de recurenta.
Avem ca \( C_n^k=C_{n-1}^{k-1}+C_{n-1}^k \)
\( S_{n}-S_{n-1}=\sum_{k=0}^{n}\displaystyle\frac{C_{n}^k}{2k+1}-\sum_{k=0}^{n-1}\frac{C_{n-1}^k}{2k+1}=\sum_{k=1}^{n}\frac{C_{n-1}^{k-1}}{2k+1}=\sum_{k=0}^{n}\frac{\frac{k}{n}C_{n}^k}{2k+1}=\frac{1}{2n}\sum_{k=0}^{n}(\frac{2k}{2k+1}C_{n}^k)=\frac{1}{2n}\sum_{k=0}^{n}(\frac{2k+1}{2k+1}-\frac{1}{2k+1})C_{n}^k=\frac{1}{2n}2^n-\frac{1}{2n}\sum_{k=0}^{n}\frac{C_n^k}{2k+1}=\frac{1}{2n}2^n-\frac{1}{2n}S_{n} \)
Rezulta ca avem \( S_{n}-S_{n-1}=\frac{1}{2n}2^n-\frac{1}{2n}S_{n} \), deci \( (2n+1)S_{n}-2nS_{n-1}=2^n \), adica \( S_{n}=\frac{2n}{2n+1}S_{n-1}+\frac{2^n}{2n+1} \)
Mai departe procedam prin inductie.
Presupunem ca \( S_{n-1}\geq\displaystyle\frac{2^n}{2n-1} \) si demonstram ca \( S_{n}\geq\displaystyle\frac{2^{n+1}}{2n+1} \)
Avem ca \( S_{n}=\frac{2n}{2n+1}S_{n-1}+\frac{2^n}{2n+1}\geq\displaystyle\frac{2n}{2n+1}\cdot\frac{2^n}{2n-1}+\frac{2^n}{2n+1}=\frac{2^n}{2n+1}(\frac{2n}{2n-1}+1)\geq\frac{2^n}{2n+1}\cdot 2=\frac{2^{n+1}}{2n+1} \), de unde rezulta ca inegalitatea este adevarata pentru orice n.