319 MATHEMATICAL COMMUNICATIONS Math Commun 16(2011), 319–324 A new refinement of the Radon inequality Cristinel Mortici1,∗ Valahia University of Tˆ argovi¸ste, Department of Mathematics, Bd Unirii 18, 130 082 Tˆ argovi¸ste, Romania Received September 1, 2009; accepted October 20, 2010 Abstract The aim of this paper is to give a new extension of the Radon inequality AMS subject classifications: 26D15 Key words: Radons inequality, Bergstră oms inequality, Hă olders inequality, Nesbitt’s inequality Introduction In this paper we study the following inequality due to J Radon, first published in [12] More precisely, for every real numbers p > 0, xk ≥ 0, ak > 0, for ≤ k ≤ n, the following inequality holds true: n xkp+1 ( ≥ apk ( k=1 p+1 n k=1 xk ) p n k=1 ak ) , p > (1) For the proof and other comments, see [12, p 1351], or the monograph [7, p 31] Inequality (1) is widely studied by many authors because of its intrinsec beauty and also because of its utility in practical applications, as in the case of obtaining more general inequalities involving manifolds [9, p 692] A particular case p = 2, x21 x2 x2 (x1 + x2 + + xn )2 + + + n ≥ , a1 a2 an a1 + a2 + + an is also well-known under the name of Bergstrăoms inequality, e.g., [3, 4, 6] It is equivalent to the Cauchy-Buniakovski-Schwarz inequality, while the Radon’s inequality (1) is equivalent to the more general Hăolders inequality Generalizations of Bergstrăom inequality and of the Cauchy-Buniakovski-Schwarz inequality are established by many authors, in particular by [1, 2, 5] and [11] and later in [8] We give here more general versions of the results from [8] and these results are also generalizations of the Radon inequality and the Hăolder inequality In this sense, note that the Radon’s inequality (1) follows from p+1 1/(p+1) 1/q n n n q x k 1/q · a ≥ xk k 1/q ak k=1 k=1 k=1 ∗ Corresponding author Email address: cmortici@valahia.ro (C Mortici) http://www.mathos.hr/mc c 2011 Department of Mathematics, University of Osijek 320 C Mortici by raising to the (p + 1)-th power, where (p + 1)−1 + q −1 = The results It is established as the main result in [8] that: x21 x2 x2 (x1 + x2 + + xn )2 (ai xj − aj xi )2 + + + n ≥ + max , 1≤i 0, ≤ k ≤ n, which is considered as an extension of the Bergstrăom inequality It was proved in [8], and before as a special case in [2, 5] and [11], that the sequence dn = x21 x2 x2 (x1 + x2 + + xn )2 + + + n − a1 a2 an a1 + a2 + + an , n ≥ 2, is increasing and inequality (2) is dn ≥ d2 , where d2 = x21 x2 (x1 + x2 )2 (a1 x2 − a2 x1 )2 + 2− = a1 a2 a1 + a2 a1 a2 (a1 + a2 ) We complete here that by putting xk = λk ak , ≤ k ≤ n, in inequality (2), a more convenient form is obtained, at least if we think that the involved maximum can be calculated more easily More precisely, a1 λ21 + a1 λ22 + + an λ2n ≥ (a1 λ1 + a2 λ2 + + an λn )2 aj (λi − λj )2 + max 1≤i 0, a, b > 0, x, y ≥ 0, xp+1 y p+1 (x + y)p+1 p(x + y)p−1 (bx − ay)2 + − ≥ ap bp (a + b)p ab(a + b)p holds Proof The expression on the left-hand side of the inequality can be written as xp+1 y p+1 (x + y)p+1 + − ap bp (a + b)p p p p x [bx(a + b)] + y [ay(a + b)] − (x + y) [ab(x + y)] = ap bp (a + b)p p x (abx + b x) − (abx + aby)p − y (abx + aby)p − (a2 y + aby)p = ap bp (a + b)p (3) Without loss of generality, we assume that bx ≥ ay As a simple consequence of the Lagrange theorem, for every p > and < u ≤ v, we have A new refinement of the Radon inequality 321 pup−1 (v − u) ≤ v p − up ≤ pv p−1 (v − u) Using this double inequality in the square brackets from (3), we obtain x (abx + b2 x)p − (abx + aby)p − y (abx + aby)p − (a2 y + aby)p ap bp (a + b)p p−1 xp(abx + aby) (b x − aby) − yp(abx + aby)p−1 (abx − a2 y) ≥ ap bp (a + b)p p−1 p(x + y) (bx − ay) = ab(a + b)p Now we are in the position to give the extension of inequality (2), which is stronger than the Radon’s inequality Theorem For every n ≥ 2, p > 0, ak > 0, xk ≥ 0, ≤ k ≤ n, it holds: xp+1 xnp+1 (x1 + x2 + + xn )p+1 xp+1 + ≥ p p + + p a1 a2 an (a1 + a2 + + an )p (xi + xj )p−1 (ai xj − aj xi )2 +p · max 1≤i 0, ≤ k ≤ n, it holds: 1 np+1 (ai − aj )2 p−1 ≥ p · · max p + p + + p − 1≤i 0, xk ≥ 0, ak > 0, ≤ k ≤ n, the following extension of the Radon’s inequality holds: xp+1 xp+1 xp+1 (x1 + x2 + + xn )p+1 n p + p + + p − a1 a2 an (a1 + a2 + + an )p p−1 (xi + xj ) (ai xj − aj xi )2 ≥ p · max 1≤i 0, ≤ k ≤ n, with s = x1 + x2 + + xn , the following extension of Nesbitt’s inequality holds: n k=1 xk ≥ p−1 · (s − xk )p s +p · n n−1 p xi xj (xi + xj )p−1 (xi − xj )2 p 1≤i