SECTION 3.3 Jensen’s Inequality 155
the objective is to show thatP(a,b)≥0 whena,b,m,nare as
given above. Regard the above as a polynomial inaand use
Descartes Rule of Signs to conclude that (counting multiplic-
ities)P(a,b) has at most four positive real zeros.
(c) Note that a = b is a zero of P(a,b), i.e., that P(b,b) = 0.
Next, show that
d
da
P(a,b)
∣∣
∣∣
∣a=b=
d^2
da^2
P(a,b)
∣∣
∣∣
∣a=b=
d^3
da^3
P(a,b)
∣∣
∣∣
∣a=b= 0.
(d) Use the above to conclude that P(a,b) ≥ 0 with equality if
and only ifa=b(orm=n).
3.3 Jensen’s Inequality
If P and Q are points in the coordinate plane, then it’s easy to see
that the set of points of the formX = X(t) = (1−t)P +tQ, 0 ≤
t ≤ 1 is precisely the line segment joining P and Q. We shall call
such a point a convex combination of P and Q. More generally,
ifP 1 , P 2 , ..., Pnare points in the plane, and if t 1 , t 2 , ..., tnare non-
negative real numbers satisfyingt 1 +t 2 +···+tn= 1, then the point
X=X(t) =t 1 P 1 +t 2 P 2 +···+tnPn
is a convex combination ofP 1 , P 2 , ..., Pn. This set is precisely smallest
convex polygon in the plane containing the pointsP 1 , P 2 , ..., Pn. Such
a polygon is depicted below: