Iloczyn z ułamkami
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mol_ksiazkowy
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Iloczyn z ułamkami
Udowodnić, że \(\displaystyle{ (1+ \frac{1}{\sqrt{1}})....(1+ \frac{1}{\sqrt{n}}) \ge 2\sqrt{n}}\).
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kerajs
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Re: Iloczyn z ułamkami
\(\displaystyle{ n=1 \ : \\
L=(1+ \frac{1}{\sqrt{1}})=2 \\
P=2 \sqrt{1} =2\\
L=P \\
\\ . \\ .
\\ n=2 \ : \\
L=(1+ \frac{1}{\sqrt{1}})(1+ \frac{1}{\sqrt{2}}) = \sqrt{2}( \sqrt{2}+1) \approx 3,414 \\
P=2 \sqrt{2}\approx 2,828\\
L>P}\)
\(\displaystyle{ Z: \ (1+ \frac{1}{\sqrt{1}})....(1+ \frac{1}{\sqrt{n}}) \ge 2\sqrt{n} \\
T: \ (1+ \frac{1}{\sqrt{1}})....(1+ \frac{1}{\sqrt{n}})(1+ \frac{1}{\sqrt{n+1}}) \ge 2\sqrt{n+1} \\
L=(1+ \frac{1}{\sqrt{1}})....(1+ \frac{1}{\sqrt{n}})(1+ \frac{1}{\sqrt{n+1}}) \ge 2\sqrt{n}(1+ \frac{1}{\sqrt{n+1}})= \frac{2}{\sqrt{n+1}} \sqrt{n }(\sqrt{n+1}+1)=\\ =
\frac{2}{\sqrt{n+1}} \sqrt{ \left[ \sqrt{n }(\sqrt{n+1}+1)\right]^2} = \frac{2}{\sqrt{n+1}} \sqrt{ n^2+2n+2n\sqrt{n+1}} >\frac{2}{\sqrt{n+1}} \sqrt{ n^2+2n+1} =\\ = \frac{2}{\sqrt{n+1}} \sqrt{ (n+1)^2}=
2\sqrt{n+1} =P
}\).
L=(1+ \frac{1}{\sqrt{1}})=2 \\
P=2 \sqrt{1} =2\\
L=P \\
\\ . \\ .
\\ n=2 \ : \\
L=(1+ \frac{1}{\sqrt{1}})(1+ \frac{1}{\sqrt{2}}) = \sqrt{2}( \sqrt{2}+1) \approx 3,414 \\
P=2 \sqrt{2}\approx 2,828\\
L>P}\)
\(\displaystyle{ Z: \ (1+ \frac{1}{\sqrt{1}})....(1+ \frac{1}{\sqrt{n}}) \ge 2\sqrt{n} \\
T: \ (1+ \frac{1}{\sqrt{1}})....(1+ \frac{1}{\sqrt{n}})(1+ \frac{1}{\sqrt{n+1}}) \ge 2\sqrt{n+1} \\
L=(1+ \frac{1}{\sqrt{1}})....(1+ \frac{1}{\sqrt{n}})(1+ \frac{1}{\sqrt{n+1}}) \ge 2\sqrt{n}(1+ \frac{1}{\sqrt{n+1}})= \frac{2}{\sqrt{n+1}} \sqrt{n }(\sqrt{n+1}+1)=\\ =
\frac{2}{\sqrt{n+1}} \sqrt{ \left[ \sqrt{n }(\sqrt{n+1}+1)\right]^2} = \frac{2}{\sqrt{n+1}} \sqrt{ n^2+2n+2n\sqrt{n+1}} >\frac{2}{\sqrt{n+1}} \sqrt{ n^2+2n+1} =\\ = \frac{2}{\sqrt{n+1}} \sqrt{ (n+1)^2}=
2\sqrt{n+1} =P
}\).
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a4karo
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Re: Iloczyn z ułamkami
Dla \(\displaystyle{ k>1}\) mamy \(\displaystyle{ \frac{1}{\sqrt{k}}>\frac{1}{\sqrt{k}+\sqrt{k-1}}=\sqrt{k}-\sqrt{k-1}}\)
Dlatego dla \(\displaystyle{ n\ge 2}\)
\(\displaystyle{ \begin{align}
\left(1+\frac{1}{\sqrt{1}}\right)&\left(1+\frac{1}{\sqrt{2}}\right)\dots\left(1+\frac{1}{\sqrt{n}}\right)\\
=2&\left(1+\frac{1}{\sqrt{2}}\right)\dots\left(1+\frac{1}{\sqrt{n}}\right)\\
>2&\left(1+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n}}\right)\\
=2&\left(1+(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\dots+(\sqrt{n}-\sqrt{n-1})\right)\\
=2&\sqrt{n}
\end{align}}\)
Dlatego dla \(\displaystyle{ n\ge 2}\)
\(\displaystyle{ \begin{align}
\left(1+\frac{1}{\sqrt{1}}\right)&\left(1+\frac{1}{\sqrt{2}}\right)\dots\left(1+\frac{1}{\sqrt{n}}\right)\\
=2&\left(1+\frac{1}{\sqrt{2}}\right)\dots\left(1+\frac{1}{\sqrt{n}}\right)\\
>2&\left(1+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n}}\right)\\
=2&\left(1+(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\dots+(\sqrt{n}-\sqrt{n-1})\right)\\
=2&\sqrt{n}
\end{align}}\)