O inegalitate pt. aplicarea multipla a derivatelor (own)
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
O inegalitate pt. aplicarea multipla a derivatelor (own)
Sa se arate ca pentru orice \( x\ge 1 \) avem \( x^x\cdot (x+1)^4\ge 16\cdot e^{2x-2} \).
Last edited by Virgil Nicula on Fri May 23, 2008 11:50 pm, edited 3 times in total.
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Bogdan Cebere
- Thales
- Posts: 145
- Joined: Sun Nov 04, 2007 1:04 pm
Fie \( f(x)=x \ln x +4 \ln (x+1) -\ln 16 -2x +2 \)
\( f^\prime=\ln x+\frac{4}{x+1}-1 \)
\( f^{\prime \prime}=\frac{1}{x}-\frac{4}{(x+1)^2}=\frac{(x-1)^2}{(x+1)^2}\geq 0 \)
Deci \( f^\prime \) este crescatoare, adica \( f^\prime(x) \geq f^\prime (1)=1 \Rightarrow \)
\( f \) crescatoare pe \( [1,\infty) \).
Atunci \( f(x) \geq f(1)=0 \) sau \( e ^ {f(x)}\geq 1 \), adica \( x^x\cdot (x+1)^4\ge 16\cdot e^{2x-2} \).
\( f^\prime=\ln x+\frac{4}{x+1}-1 \)
\( f^{\prime \prime}=\frac{1}{x}-\frac{4}{(x+1)^2}=\frac{(x-1)^2}{(x+1)^2}\geq 0 \)
Deci \( f^\prime \) este crescatoare, adica \( f^\prime(x) \geq f^\prime (1)=1 \Rightarrow \)
\( f \) crescatoare pe \( [1,\infty) \).
Atunci \( f(x) \geq f(1)=0 \) sau \( e ^ {f(x)}\geq 1 \), adica \( x^x\cdot (x+1)^4\ge 16\cdot e^{2x-2} \).