Consider a set of n positive real numbers 
, with 
for all i=1,2,…,n. Show then that the inequality
\ge1-\sum_{i=1}^nx_i)
holds.
Our inequality is
(1-x_2)\cdots(1-x_n)\ge1-x_1-x_2-\cdots-x_n)
.
For n=1, both sides of the above reduce to 
, and thus we have equality.
Now, suppose that the inequality is true for a set of n numbers , and let us choose any )
. Then if we multiply both sides of the inequality by 
, we get
\prod_{i=1}^n(1-x_i)\ge\left(1-\sum_{i=1}^nx_i\right)(1-x_{n+1}))
\ge\left(1-\sum_{i=1}^nx_i\right)(1-x_{n+1}))
.
Now, expanding on the right hand side:
(1-x_{n+1})&=&1-\sum_{i=1}^nx_i-x_{n+1}+x_{n+1}\sum_{i=1}^nx_i\\&=&1-\sum_{i=1}^{n+1}x_i+x_{n+1}\sum_{i=1}^nx_i\\&\ge&1-\sum_{i=1}^{n+1}x_i\end{eqnarray})
since 
. Thus the equality holds for the set 
, and so, considering the base case of n=1, we have the inequality established by induction.
For a more explicit exemplar of this construction, we apply the above process to the base case:
&=&(1-x_1)\\(1-x_1)(1-x_2)&=&(1-x_1)(1-x_2)\\(1-x_1)(1-x_2)&=&1-x_1-x_2+x_1x_2\\(1-x_1)(1-x_2)&\ge&1-x_1-x_2\end{eqnarray})
,
since 
.
Next,
(1-x_2)&\ge&1-x_1-x_2\\(1-x_1)(1-x_2)(1-x_3)&\ge&(1-x_1-x_2)(1-x_3)\\(1-x_1)(1-x_2)(1-x_3)&\ge&1-x_1-x_2-x_3+(x_1+x_2)x_3\\(1-x_1)(1-x_2)(1-x_3)&\ge&1-x_1-x_2-x_3\\\end{eqnarray})
since x_3\gt{0})
.
And so on.
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Tags: Inequality, Math, Monday Math
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