# 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(\(\frac{z+x}{2}\)))

ProofWithout 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 ≤ \(\frac{y+z}{2}\) ≤ \(\frac{x+y+z}{3}\)

NOTEwhen 'f' is a concave function then the INEQUALITY Gets Reversed.

**Cite as:**Popoviciu's inequality.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/popovicius-inequality/