Toán Lớp 10: Cho a, b >0 chứng minh rằng: (1+a/b)^n+(1+b/a)^n >= 2^(n+1)
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Toán Lớp 10: Cho a, b >0 chứng minh rằng: (1+a/b)^n+(1+b/a)^n >= 2^(n+1)
Comments ( 1 )
\text{Áp dụng bất đẳng thức AM-GM ta được:}\\
\left(1 + \dfrac ab\right)^n \geqslant \left(2\sqrt{\dfrac ab}\right)^n = 2^n\cdot \left(\sqrt{\dfrac ab}\right)^n\\
\left(1 + \dfrac ba\right)^n \geqslant \left(2\sqrt{\dfrac ba}\right)^n = 2^n\cdot \left(\sqrt{\dfrac ba}\right)^n\\
\text{Cộng vế theo vế ta được:}\\
\left(1 + \dfrac ab\right)^n+\left(1 + \dfrac ba\right)^n \geqslant 2^n\cdot \left(\sqrt{\dfrac ab}\right)^n+2^n\cdot \left(\sqrt{\dfrac ba}\right)^n\\
\Leftrightarrow \left(1 + \dfrac ab\right)^n+\left(1 + \dfrac ba\right)^n \geqslant 2^n\cdot \left[\left(\sqrt{\dfrac ab}\right)^n+\left(\sqrt{\dfrac ba}\right)^n\right]\\
\text{Ta lại có:}\\
\left(\sqrt{\dfrac ab}\right)^n+\left(\sqrt{\dfrac ba}\right)^n\geqslant 2\cdot\sqrt{\left(\sqrt{\dfrac ab}\right)^n\cdot\left(\sqrt{\dfrac ba}\right)^n} =2\\
\text{Do đó:}\\
\left(1 + \dfrac ab\right)^n+\left(1 + \dfrac ba\right)^n \geqslant 2^n\cdot 2 = 2^{n+1}\\
\text{Dấu = xảy ra khi và chỉ khi}\\
\begin{cases}1 = \dfrac ab\\1 = \dfrac ba\\ \sqrt{\dfrac ab}=\sqrt{\dfrac ba}\end{cases}\Longleftrightarrow a = b
\end{array}\)