The inequality below has been discovered by J. Radon in 1913 but is often referred to, along with Bergström's inequality (of which it is a generalization) as Titu's Lemma or Titu's inequality in honor of Titu Andreescu after the publication of Mathematical Olympiad Treasures in 2003.

The content of the present page has been borrowed (at least in its initial form) from an article by Dorin Marghidanu Generalizations and Refinements for Bergström's and Radon's Inequalities.

Clearly, for $p=1\;$ the inequality becomes that of Bergström.

As a first step, we prove the inequality for $n=2,\;$ deriving it from the well-known Hölder's inequality:

$\displaystyle\sum_{i=1}^nu_iv_i\le\left(\sum_{i=1}^nu_i^s\right)^{1/s}\left(\sum_{i=1}^nv_i^t\right)^{1/t},$

where $\displaystyle \frac{1}{s}+\frac{1}{t}=1,\;$ $s,t\gt 1,\;$ and all $u_i\;$ and $v_i\;$ are assumed positive. This is obviously a generalization of the Cauchy-Schwarz inequality The same method will also work for larger $n\;$ but I prefer to use Dorin Marghidanu's original derivation that depends on the case of $n=2.$

Thus, we want to prove that, say,

$\displaystyle\frac{x^{p+1}}{a^p}+\frac{y^{p+1}}{b^p}\ge\frac{(x+y)^{p+1}}{(a+b)^p}.$

Setting $\displaystyle s=\frac{p+1}{p}\;$ and $t=p+1,\;$ We start with

\displaystyle\begin{align} x+y &= a^{1/s}\left(\frac{x}{a^{1/s}}\right)+b^{1/s}\left(\frac{y}{b^{1/s}}\right)\\ &\le (a^{s/s}+b^{s/s})^{1/s}\left(\frac{x^t}{a^{t/s}}+\frac{y^t}{b^{t/s}}\right)^{1/t}\\ &=\left[(a+b)^{p}\left(\frac{x^{p+1}}{a^{p}}+\frac{y^{p+1}}{b^{p}}\right)\right]^{1/{(p+1)}}. \end{align}

This is equivalent to the required inequality. Now for the rest of $n.\;$ Define

$\displaystyle d_n=\frac{x_1^{p+1}}{a_1^p}+\frac{x_2^{p+1}}{a_2^p}+\cdots+\frac{x_n^{p+1}}{a_n^p}-\frac{(x_1+x_2+\cdots+x_n)^{p+1}}{a_1+a_2+\cdots+a_n}^p.$

Our task is to prove that $d_n\ge 0,\;$ for $n\ge 2.\;$ We are going to show more, viz., that the sequence $\{d_n\}\;$ monotone increasing and, since $d_1=0,\;$ this will solve the entire problem of proving Radon's inequality.

To this end,

\displaystyle\begin{align} d_{n+1}-d_{n} &= \sum_{k=1}^{n+1}\frac{x_k^{p+1}}{a_k^p}-\frac{\left(\displaystyle\sum_{k=1}^{n+1}x_k \right)^{p+1}}{\left(\displaystyle\sum_{k=1}^{n+1}a_k \right)^{p}} -\sum_{k=1}^{n}\frac{x_k^{p+1}}{a_k^p}+\frac{\left(\displaystyle\sum_{k=1}^{n}x_k \right)^{p+1}}{\left(\displaystyle\sum_{k=1}^{n}a_k \right)^{p}}\\ &=\left[\frac{\left(\displaystyle\sum_{k=1}^{n}x_k \right)^{p+1}}{\left(\displaystyle\sum_{k=1}^{n}a_k \right)^{p}}+\frac{x_{n+1}^{p+1}}{a_{n+1}^p}\right]-\frac{\left(\displaystyle\sum_{k=1}^{n+1}x_k \right)^{p+1}}{\left(\displaystyle\sum_{k=1}^{n+1}a_k \right)^{p}}\\ &\ge \frac{\left(\displaystyle\sum_{k=1}^{n+1}x_k \right)^{p+1}}{\left(\displaystyle\sum_{k=1}^{n+1}a_k \right)^{p}}-\frac{\left(\displaystyle\sum_{k=1}^{n+1}x_k \right)^{p+1}}{\left(\displaystyle\sum_{k=1}^{n+1}a_k \right)^{p}}\\ &=0, \end{align}

where in the penultimate step we used the earlier case of $n=2.$

Obviously, this proof can be regarded as a proof by induction.

Dan Sitaru has kindly alerted me to the validity of what's known as the reverse Radon's inequality:

If $x_k, a_k \gt 0,\;$ $k\in\{1,2,\ldots,n\},\;$ $0\le p\le 1,\;$ then

$\displaystyle \frac{x_1^{p}}{a_1^{p-1}}+\frac{x_2^{p}}{a_2^{p-1}}+\cdots+\frac{x_n^{p}}{a_n^{p-1}}\le\frac{(x_1+x_2+\cdots+x_n)^{p}}{(a_1+a_2+\cdots+a_n)^{p-1}}.$

### Applications

1. A Problem in Four Variables

Dan Sitaru has posted the following problem from the Romanian Mathematical Magazine:

If $a,b,c,d\in (0,\infty ),\;$ and $abcd=1\;$ then

$\displaystyle\frac{(a+b+c)^5}{(b+c+d)^4}+\frac{(b+c+d)^5}{(c+d+a)^4}+\frac{(c+d+a)^5}{(d+a+b)^4}+\frac{(d+a+b)^5}{(a+b+c)^4}\ge 12.$

The inequality is solved by an application of Radon's inequality, followed by the AM-GM inequality:

\displaystyle\begin{align} \frac{(a+b+c)^5}{(b+c+d)^4}&+\frac{(b+c+d)^5}{(c+d+a)^4}+\frac{(c+d+a)^5}{(d+a+b)^4}+\frac{(d+a+b)^5}{(a+b+c)^4} \\ &\ge\frac{[3(a+b+c+d)]^5}{[3(a+b+c+d)]^4}\\ &=3(a+b+c+d)\\ &\ge 3\cdot 4(abcd)^{1/4}\\ &\ge 12. \end{align}

2. 42 IMO, Problem 2

Prove that, for all positive $a,b,c,$

$\displaystyle\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}\ge 1.$

The left-hand side can be rewritten as

$\displaystyle M=\frac{a^{3/2}}{\sqrt{a^3+8abc}}+\frac{b^{3/2}}{\sqrt{b^3+8abc}}+\frac{c^{3/2}}{\sqrt{c^3+8abc}}$

which suggests using Radon's inequality with $\displaystyle p=\frac{1}{2}\;$ and $n=3:$

$\displaystyle M\ge\frac{(a+b+c)^{3/2}}{(a^3+b^3+c^3+24abc)^{1/2}}=\sqrt{\frac{(a+b+c)^{3}}{a^3+b^3+c^3+24abc}}$

Thus suffice it to prove that $\displaystyle\frac{(a+b+c)^{3}}{a^3+b^3+c^3+24abc}\ge 1.\;$ This inequality reduces to

$ab^2+a^2b+bc^2+b^2c+ca^2+c^2a\ge 6abc$

which is an immediate consequence of the AM-GM inequality.

3. Dorin Marghidanu's Example

Three solutions can be found on a separate page.