Sumy i iloczyny
: 22 lip 2023, o 18:31
Udowodnić, że \(\displaystyle{ \sum_{i=1}^{n} |a_i-b_i| \geq | \prod_{1=1}^{n} a_i - \prod_{1=1}^{n} b_i|}\) dla liczb \(\displaystyle{ a_i, b_i }\) z przedziału \(\displaystyle{ [-1,1]}\).
ZIM:
Dla \(\displaystyle{ n=1}\) jest ok. Niech \(\displaystyle{ n\in\NN}\) będzie takie, że
\(\displaystyle{ \left| \prod_{i=1}^{n}a_i - \prod_{i=1}^{n}b_i \right| \le \sum_{i=1}^{n} |a_i-b_i|. }\)
Jako, że
\(\displaystyle{
\begin{split}
\prod_{i=1}^{n+1} a_i - \prod_{i=1}^{n+1} b_i & = a_{n+1} \prod_{i=1}^{n} a_i - b_{n+1} \prod_{i=1}^{n}b_i \\[1.5ex]
&=a_{n+1} \prod_{i=1}^{n} a_i \red{- b_{n+1} \prod_{i=1}^{n}a_i + b_{n+1} \prod_{i=1}^{n}a_i } - b_{n+1} \prod_{i=1}^{n}b_i \\[1.5ex]
& =\left( a_{n+1} - b_{n+1}\right) \prod_{i=1}^{n}a_i + b_{n+1} \left( \prod_{i=1}^{n}a_i - \prod_{i=1}^{n}b_i \right).
\end{split} }\)
To też
\begin{split}
\prod_{i=1}^{n+1} a_i - \prod_{i=1}^{n+1} b_i & = a_{n+1} \prod_{i=1}^{n} a_i - b_{n+1} \prod_{i=1}^{n}b_i \\[1.5ex]
&=a_{n+1} \prod_{i=1}^{n} a_i \red{- b_{n+1} \prod_{i=1}^{n}a_i + b_{n+1} \prod_{i=1}^{n}a_i } - b_{n+1} \prod_{i=1}^{n}b_i \\[1.5ex]
& =\left( a_{n+1} - b_{n+1}\right) \prod_{i=1}^{n}a_i + b_{n+1} \left( \prod_{i=1}^{n}a_i - \prod_{i=1}^{n}b_i \right).
\end{split} }\)
\(\displaystyle{
\begin{split}
\left| \prod_{i=1}^{n+1} a_i - \prod_{i=1}^{n+1} b_i \right| & \le \left|a_{n+1} - b_{n+1}\right| \left|\prod_{i=1}^{n} a_i\right| + \left| b_{n+1}\right| \left| \prod_{i=1}^{n}a_i - \prod_{i=1}^{n}b_i \right|\\[1.5ex]
&\le \left|a_{n+1} - b_{n+1}\right| + \left| \prod_{i=1}^{n}a_i - \prod_{i=1}^{n}b_i \right|\\[1.5ex]
&\le \left|a_{n+1} - b_{n+1}\right| +\sum_{i=1}^{n} |a_i-b_i|\\[1.5ex]
&=\sum_{i=1}^{n+1} |a_i-b_i|.
\end{split}
}\)
ZIM \(\displaystyle{ \blacksquare}\)\begin{split}
\left| \prod_{i=1}^{n+1} a_i - \prod_{i=1}^{n+1} b_i \right| & \le \left|a_{n+1} - b_{n+1}\right| \left|\prod_{i=1}^{n} a_i\right| + \left| b_{n+1}\right| \left| \prod_{i=1}^{n}a_i - \prod_{i=1}^{n}b_i \right|\\[1.5ex]
&\le \left|a_{n+1} - b_{n+1}\right| + \left| \prod_{i=1}^{n}a_i - \prod_{i=1}^{n}b_i \right|\\[1.5ex]
&\le \left|a_{n+1} - b_{n+1}\right| +\sum_{i=1}^{n} |a_i-b_i|\\[1.5ex]
&=\sum_{i=1}^{n+1} |a_i-b_i|.
\end{split}
}\)