Difference between revisions of "Radon's Inequality"
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<cmath> \frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+1} } { (b_1 + b_2 + \cdots+ b_n )^p} </cmath> | <cmath> \frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+1} } { (b_1 + b_2 + \cdots+ b_n )^p} </cmath> | ||
| − | It is a direct consequence of [[Hölder's Inequality]], and a generalization of [[Titu's Lemma]]. | + | It is a direct consequence of [[Hölder's Inequality]], and a generalization of [[Titu's Lemma]] (for p=1, it is just that). |
=== Proof === | === Proof === | ||
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Just apply Hölder for: | Just apply Hölder for: | ||
| − | <cmath>(b_1 + b_2 + \cdots+ b_n )^{\frac{p}{p+1}}\ | + | <cmath>(b_1 + b_2 + \cdots+ b_n )^{p/(p+1)}\left(\frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p }\right)^{1/(p+1)} \geq a_1 + a_2 + \cdots+ a_n \Leftrightarrow</cmath> |
| + | <cmath> \frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+1} } { (b_1 + b_2 + \cdots+ b_n )^p} </cmath> | ||
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| + | === Further Generalizations === | ||
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| + | <cmath> \frac{ a_1^{p+m} } { b_1^p } + \frac{ a_2 ^{p+m} } { b_2^p } + \cdots + \frac{ a_n ^{p+m} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+m} } { (b_1 + b_2 + \cdots+ b_n )^p} </cmath> | ||
Latest revision as of 08:00, 11 August 2024
Radon's Inequality states:
![\[\frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+1} } { (b_1 + b_2 + \cdots+ b_n )^p}\]](http://latex.artofproblemsolving.com/5/b/7/5b752a334e82327fa3d75451b5a77efffa369081.png)
It is a direct consequence of Hölder's Inequality, and a generalization of Titu's Lemma (for p=1, it is just that).
Proof
Just apply Hölder for:
![\[(b_1 + b_2 + \cdots+ b_n )^{p/(p+1)}\left(\frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p }\right)^{1/(p+1)} \geq a_1 + a_2 + \cdots+ a_n \Leftrightarrow\]](http://latex.artofproblemsolving.com/e/4/a/e4a17834c130bf871ff85452d9b2244a9baaa348.png)
Further Generalizations
![\[\frac{ a_1^{p+m} } { b_1^p } + \frac{ a_2 ^{p+m} } { b_2^p } + \cdots + \frac{ a_n ^{p+m} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+m} } { (b_1 + b_2 + \cdots+ b_n )^p}\]](http://latex.artofproblemsolving.com/f/d/8/fd856b700217375a146e69001160947c8ee8c62c.png)
