sorry calculation mistake in second last step..
66That's basically the power mean inequality.
Evaluate \lim_{x->0}\left[ \frac{a^x+b^x+c^x}{3}\right]^{1/x}
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4 Answers
1\lim_{x \to 0}(1+f(x))^{\frac{1}{x}}=e^{f'(x)} \ \ if \lim_{x \to 0}(f(x))=0\\ here \ f(x)=\frac{a^x+b^x+c^x}{3}-1 \\ \texttt{hence limit is } e^{\frac{\ln a + \ln b + \ln c}{3}}=(abc)^{\frac{1}{3}}
62The proof I had in mind was slightly different.
Since x is close to zero,
ax≈bx=cx
Hence applying AM GM inequality, we get
\frac{a^x+b^x+c^x}{3}\geq \left(a^x\times b^x \times c^x \right)^{1/3}=(abc)^{x/3}
Now take the 1/x power on both sides to get
\left[ \frac{a^x+b^x+c^x}{3}\right]^{1/x}\geq(abc)^{1/3}
since the three terms are equal, equality will hold hence the answer will be 61/3