Radon's Inequality

Radon's Inequality states:

$\frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+1} } { (b_1 + b_2 + \cdots+ b_n )^p}$

It is a direct consequence of Hölder's Inequality, and a generalization of Titu's Lemma (for p=2, it is just that).

Proof

Just apply Hölder for:

$(b_1 + b_2 + \cdots+ b_n )^{p/(p+1)}\left(\frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p }\right)^{1/(p+1)} \geq a_1 + a_2 + \cdots+ a_n \Leftrightarrow$ $\frac{ a_1^{p+1} } { b_1^p } + \frac{ a_2 ^{p+1} } { b_2^p } + \cdots + \frac{ a_n ^{p+1} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+1} } { (b_1 + b_2 + \cdots+ b_n )^p}$

Further Generalizations

$\frac{ a_1^{p+m} } { b_1^p } + \frac{ a_2 ^{p+m} } { b_2^p } + \cdots + \frac{ a_n ^{p+m} } { b_n^p } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^{p+m} } { (b_1 + b_2 + \cdots+ b_n )^p}$

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Radon's Inequality

Proof

Further Generalizations