Fie \( a \) şi \( b \) două numere reale strict pozitive.
a) Demonstraţi că, dacă \( n\in\mathbf{N} \), \( \left(1+\frac{a}{b}\right)^n+\left(1+\frac{b}{a}\right)^n\geq 2^{n+1} \)
b) Demonstraţi aceeaşi inegalitate când \( n\in\mathbf{Z} \).
Test pentru classe preparatoire, 2008, problema 2
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andy crisan
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Se demonstreaza usor prin inductie ca \( \frac{a^{n}+b^{n}}{2}\geq \left(\frac{a+b}{2}\right)^{n} \)\( \forall n \in \mathbb{N}^{*} \).Cu \( a>0, b>0 \) cu egalitatea daca si numai daca \( a=b \) sau \( n=1 \). Folosind aceasta inegalitate in enunt avem ca \( \left(1+\frac{a}{b}\right)^{n}+\left(1+\frac{b}{a}\right)^{n}\geq \left(\frac{2+\frac{a}{b}+\frac{b}{a}}{2}\right)^{n}\geq 2^{n+1}. \)
La b) consideram \( n<0 \) si \( m=-n \) si folosind aceeasi inegalitate de la punctul a) avem ca \( LHS= \frac{1}{\left(1+\frac{a}{b}\right)^{m}}+ \frac{1}{\left(1+\frac{b}{a}\right)^{m}} =\left(\frac{a}{a+b}\right)^{m}+\left(\frac{b}{a+b}\right)^{m}\geq \left(\frac{\frac{a}{a+b}+\frac{b}{a+b}}{2}\right)^{m}\cdot 2 =\frac{1}{2^{m}} \cdot 2=\frac{1}{2^{m-1}}. \)
La b) consideram \( n<0 \) si \( m=-n \) si folosind aceeasi inegalitate de la punctul a) avem ca \( LHS= \frac{1}{\left(1+\frac{a}{b}\right)^{m}}+ \frac{1}{\left(1+\frac{b}{a}\right)^{m}} =\left(\frac{a}{a+b}\right)^{m}+\left(\frac{b}{a+b}\right)^{m}\geq \left(\frac{\frac{a}{a+b}+\frac{b}{a+b}}{2}\right)^{m}\cdot 2 =\frac{1}{2^{m}} \cdot 2=\frac{1}{2^{m-1}}. \)