Daca \( a>1 \) , atunci aratati ca :
\( log_{a}(a+n) \le log_{a}^n(1+a)\ ,\ (\forall) x \in N \).
prof. Mihai Dicu (generalizarea unei probleme din G.M.1/2007)
Inegalitate Logaritmica
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Mateescu Constantin
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Demonstratie : Aratam prin inductie \( p(n)\ :\ \log_a(a+n)\ \le\ \log_a^n(1+a)\ ,\ \forall\ n\ \in\ \mathbb{N} \) .Al3xx wrote:Daca \( a\ >\ 1 \) aratati ca : \( \log_a(a+n)\ \le\ \log_a^n(1+a)\ ,\ \forall\ n\ \in\ \mathbb N \) .
Pentru \( n=0\ \Longrightarrow\ \log_aa\ \le\ 1 \) , evident adevarat .
Sa presupunem \( p(k) \) adevarata pana la un \( k \) oarecare, \( k \) fixat , \( k\ge 1\ : \)
\( \Longrightarrow\ \log_a(a+k)\ \le\ \log_a^k(1+a)\ \Longrightarrow\ a+k\ \le\ a^{\log_a^k(1+a)}\ (\ast) \) .
Sa demonstram ca \( p(k+1)\ :\ \log_a(a+k+1)\ \le\ \log_a^{k+1}(1+a) \) este adevarata .
Avem \( a^{\log_a^{k+1}(1+a)}=\[a^{\log_a(1+a)}\]^{\log_a^k(1+a)}=(1+a)^{\log_a^k(1+a)}\ \ge\ 1+a^{\log_a^k(1+a)}\ \ge^{(\ast)}\ 1+a+k \) .
Deci , \( \log_a^{k+1}(1+a)\ \ge\ \log_a(1+a+k) \) adica \( p(k+1) \) este adevarata . Prin urmare problema este rezolvata .
Cazuri particulare : \( \left\|\ \begin{array}{cc}
a=e & \Longrightarrow & \ln(e+n) & \le & \ln^n(e+1) \\\\\\\\
a=10 & \Longrightarrow & \lg(10+n) & \le & \lg^n\ 11\ \end{array}\ \right\| \)
Last edited by Mateescu Constantin on Mon Feb 22, 2010 10:49 am, edited 1 time in total.
Metoda II
\( a>1\ \Longleftrightarrow \) functia logaritmica de baza \( a \) este strict crescatoare.
Aplicand Bernoulli de 2 ori \( \Longleftrightarrow \)
\( \Longleftrightarrow\ log_{a}^n(1+a)\ =\ log_{a}^n\[a\(1+\frac{1}{a}\)\]\ =\ \(1 + log_{a}\(1+\frac{1}{a}\)\)^n\ \ge\ 1+nlog_{a}\(1 + \frac{1}{a}\)\ =\ 1+log_{a}\(1+\frac{1}{a}\)^n\ \ge\ 1+log_{a}\(1+\frac{n}{a}\)\ =\ log_{a}(n+a) \)
\( a>1\ \Longleftrightarrow \) functia logaritmica de baza \( a \) este strict crescatoare.
Aplicand Bernoulli de 2 ori \( \Longleftrightarrow \)
\( \Longleftrightarrow\ log_{a}^n(1+a)\ =\ log_{a}^n\[a\(1+\frac{1}{a}\)\]\ =\ \(1 + log_{a}\(1+\frac{1}{a}\)\)^n\ \ge\ 1+nlog_{a}\(1 + \frac{1}{a}\)\ =\ 1+log_{a}\(1+\frac{1}{a}\)^n\ \ge\ 1+log_{a}\(1+\frac{n}{a}\)\ =\ log_{a}(n+a) \)