CONVEXITY ACCORDING TO THE GEOMETRIC MEAN 1 ...

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162 CONSTANTIN P. NICULESCUfor every multiplicatively convex function f which is continuous on [0 1):We shall give another application of Proposition 3.1, which seems to be new evenfor polynomials with non-negative coefficients:THEOREM 3.3. (The multiplicative analogue of Popoviciu’s Inequality [11]) : Supposethat f : I ! (0 1) is a multiplicatively convex function. Thenf (x) f (y) f (z) f 3 ( 3p xyz) > f 2 ( p xy) f 2 ( p yz) f 2 ;p zx for every x y z 2 I: Moreover, for the strictly multiplicatively convex functions theequality occurs only when x = y = z:Proof. Without loss of generality we may assume that x > y > z: Thenp xy >p zx >p yz and x >3 p xyz > z:If x > 3p xyz > y > z the desired conclusion follows from Proposition 3.1applied tox 1 = x x 2 = x 3 = x 4 = 3p xyz x 5 = y x 6 = zy 1 = y p 2 = xy y 3 = y p 4 = xz y 5 = y p 6 = yzwhile in the case x > y > 3p xyz > z we have to considerx 1 = x x 2 = y x 3 = x 4 = x 5 = 3p xyz x 6 = zy 1 = y 2 = p xy y 3 = y 4 = p xz y 5 = y 6 = p yz : According to Theorem 3.3 (applied to f (x) =e x ) , for every x y z > 0wehavex + y + z3+ 3p xyz > 2 3; pxy+p yz +p zxunless x = y = z:Other homogeneous inequalities can be obtained by extending Proposition 3.3 tolonger sequences and/or to more general convex combinations.4. Multiplicative convexity of special functionsWe start this section by recalling the following result:PROPOSITION 4.1. (P. Montel [10]) Let f : [0 a) ! [0 1) be a continuousfunction, which is multiplicatively convex on (0 a) . ThenZ xF(x) =0f (t) dtis also continuous on [0 a) and multiplicatively convex on (0 a) .
CONVEXITY ACCORDING TO THE GEOMETRIC MEAN 163Proof. Montel’s original argument was based on the fact that under the presenceof continuity, f is multiplicatively convex if, and only if,2f (x) 6 k α f (kx) +k ;α f (x=k)for every x 2 I and every k > 0 such that kx and x=k both belong to I:Actually, due to the continuity of F it suffices to show that(F( p xy)) 2 6 F(x) F(y) for every x y 2 [0 a)which is a consequence of the corresponding inequality at the level of integral sums," pxy Xn;1 p xy# 2 "x Xn;1f k 6 f k x #" y Xn;1f k y #nn n n n ni.e., ofk = 0" n;1Xk = 0fkp xyn# 2 "X n;16k = 0k = 0To see that the latter inequality holds, notice that p xy 2 h f k 6 f k x nnand then apply the Cauchy-Schwarz Inequality.fk = 0k x #" Xn;1f k y nn # :k = 0ihfk y inAs tan is continuous on [0 π=2) and multiplicatively convex on (0 π=2) arepeated application of Proposition 4.1 shows us that the Lobacevski’s function,Z xL(x) =;0log cos tdtis multiplicatively convex on (0 π=2) .tStarting withsin tand then switching to sin tt which is multiplicatively concave,a similar argument leads us to the fact that the integral sine,Z xsin tSi(x) = dt tis multiplicatively concave on (0 π=2):Another striking example is the following:0PROPOSITION 4.2. Γ is a strictly multiplicatively convex function on [1 1):Proof. In fact, log Γ(1 + x) is strictly convex and increasing on (1 1) .Or,anincreasing strictly convex function of a strictly convex function is strictly convex too.So, F(x) =log Γ(1 + e x ) is strictly convex on (0 1) and thusF(log x)Γ(1 + x) =e
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