Popoviciu's inequality
Popoviciu’s inequality, which will be used in the same manner as Jensen’s inequality. But we must note that this inequality is stronger then Jensen’s inequality, i.e. in some cases this inequality can be a powerful tool for proving other inequalities, where Jensen’s inequality does not work.
Popoviciu’s INEQUALITY
Let f : [a, b]?R be a convex function on the interval [a, b]. Then for any x, y, z ? [a, b] we have
f(\(\frac{x+y+z}{3}\)) + \(\frac{f(x) + f(y) + f(z)}{3}\) ≥ \(\frac{2}{3}\)(f(\(\frac{x+y}{2}\)) + f(\(\frac{y+z}{2}\)) + f(\(\frac{z+x}{2}\)))
Proof
Without loss of generality we assume that x ≤ y ≤ z .
If y ≤ \(\frac{x+y+z}{3}\) then
and
If x ≤ \(\frac{x+y+z}{3}\) then
\(\frac{x+y+z}{3}\) ≤ \(\frac{x+z}{2}\) ≤ z
and
\(\frac{x+y+z}{3}\) ≤ \(\frac{y+z}{2}\) ≤ z.
Therefore there exist s, t ∈ [0, 1] such that
\(\frac{x+z}{2}\) = (\(\frac{x+y+z}{3}\))s + z(1-s) ----------------(A)
\(\frac{y+z}{2}\) = (\(\frac{x+y+z}{3}\))t + z(1-t) ----------------(B)
on adding (A) and (B) :-
\(\frac{x+y-2z}{2}\) = \(\frac{x+y-2z}{3}\)(s+t)
(s+t) = \(\frac{3}{2}\)
as f is a convex function :-
f(\(\frac{x+z}{2}\)) ≤ s . f(\(\frac{x+y+z}{3}\)) + (1-s).f(z) ,
f(\(\frac{y+z}{2}\)) ≤ t . f(\(\frac{x+y+z}{3}\)) + (1-t).f(z)
and
f(\(\frac{x+y}{2}\)) ≤ \(\frac{1}{2}\).f(x) + \(\frac{1}{2}\).f(y) .
After adding together the last three inequalities we obtain the required inequality.
the Case When \(\frac{x+y+z}{3}\) ≤ y is considered similarly, bearing in mind that x ≤ \(\frac{x+z}{2}\) ≤ \(\frac{x+y+z}{3}\) and x ≤ \(\frac{y+z}{2}\) ≤ \(\frac{x+y+z}{3}\)
NOTE
when 'f' is a concave function then the INEQUALITY Gets Reversed.