530 Real Analysis
which is, in fact, the condition forfto be convex.
(P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics, Springer, 2004)
439.The functionf(t)=sintis concave on the interval[ 0 ,π]. Jensen’s inequality
yields
sinA+sinB+sinC≥3 sinA+B+C
3
=3 sinπ
3=
3
√
3
2
.
440.If we setyi=lnxi, thenxi∈( 0 , 1 ]impliesyi≤0,i= 1 , 2 ,...,n. Consider the
functionf:(−∞, 0 ]→R,f(y)=( 1 +ey)−^1. This function is twice differentiable and
f′′(y)=ey(ey− 1 )( 1 +ey)−^3 ≤ 0 , fory≤ 0.It follows that this function is concave, and we can apply Jensen’s inequality to the points
y 1 ,y 2 ,...,ynand the weightsa 1 ,a 2 ,...,an. We have
∑ni= 1ai
1 +xi=
∑ni= 1ai
1 +eyi≤
1
1 +e∑n
i= 1 aiyi=1
1 +
∏n
i= 1 eaiyi=
1
1 +
∏n
i= 1 xai
i,
which is the desired inequality.
(D. Bu ̧sneag, I. Maftei,Teme pentru cercurile ̧si concursurile de matematic ̆a(Themes
for mathematics circles and contests), Scrisul Românesc, Craiova)
441.First solution: Apply Jensen’s inequality to the convex functionf(x)=x^2 and to
x 1 =a 12 +a 22 +a^23
2 a 2 a 3,x 2 =a 12 +a 22 +a^23
2 a 3 a 1,x 3 =a^21 +a 22 +a^23
2 a 1 a 2,
λ 1 =a 12
a 12 +a 22 +a^23,λ 2 =a 22
a 12 +a 22 +a^23,λ 3 =a 32
a^21 +a 22 +a^23.
The inequality
f(λ 1 x 2 +λ 2 x 2 +λ 3 x 3 )≤λ 1 f(x 1 )+λ 2 f(x 2 )+λ 3 f(x 3 )translates to
(a 13 +a^32 +a 33 )^2
4 a 12 a^22 a^23≤
(a^41 +a 24 +a^43 )(a 12 +a^22 +a 32 )
4 a 12 a^22 a^23,
and the conclusion follows.
Second solution: The inequality from the statement is equivalent to