Difference between revisions of "Callebaut's Inequality"
Spanferkel (talk | contribs) (Created page with ''''Callebaut's Inequality''' states that for <math>1\ge x\ge y\ge 0,</math> <cmath> \sum_{i=1}^{n}a_{i}^{1+x}b_{i}^{1-x}\sum_{i=1}^{n}a_{i}^{1-x}b_{i}^{1+x}\ge\sum_{i=1}^{n}a_{i}…') |
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\[ | \[ | ||
| − | f(1+y)^{\frac xy-1}\cdot f(1-y)^{\frac xy-1} | + | f(1+y)^{\frac xy-1}\cdot f(1-y)^{\frac xy-1} \ge f(1)^{2(\frac xy-1)},\\ \\ |
| − | f(1-x)\cdot f(1)^{\frac xy-1} | + | f(1-x)\cdot f(1)^{\frac xy-1} \ge f(1-y)^{\frac xy},\\ \\ |
| − | f(1+x)\cdot f(1)^{\frac xy-1} | + | f(1+x)\cdot f(1)^{\frac xy-1} \ge f(1+y)^{\frac xy},\\ \\ |
\] | \] | ||
Multiplying the last three lines yields <math>f(1+x)f(1-x)\ge f(1+y)f(1-y)</math> as required. | Multiplying the last three lines yields <math>f(1+x)f(1-x)\ge f(1+y)f(1-y)</math> as required. | ||
Revision as of 16:09, 19 September 2010
Callebaut's Inequality states that for 
![\[\sum_{i=1}^{n}a_{i}^{1+x}b_{i}^{1-x}\sum_{i=1}^{n}a_{i}^{1-x}b_{i}^{1+x}\ge\sum_{i=1}^{n}a_{i}^{1+y}b_{i}^{1-y}\sum_{i=1}^{n}a_{i}^{1-y}b_{i}^{1+y}.\]](http://latex.artofproblemsolving.com/2/8/d/28d64c5e319b182d5064c2cb7b6c595b12749ced.png)
It can be considered as an interpolation or a refinement of Cauchy-Schwarz, which is the 
case.
Proof
Let 
. Then by Hölder,
, further (because 
)
and
.
Raising these three respectively to the 
th, 
th, th power, we get
\[ f(1+y)^{\frac xy-1}\cdot f(1-y)^{\frac xy-1} \ge f(1)^{2(\frac xy-1)},\\ \\ f(1-x)\cdot f(1)^{\frac xy-1} \ge f(1-y)^{\frac xy},\\ \\ f(1+x)\cdot f(1)^{\frac xy-1} \ge f(1+y)^{\frac xy},\\ \\ \]
Multiplying the last three lines yields 
as required.
