160 CHAPTER 3 Inequalities
Indeed, from the above, we have by settingr=
q
p
>1 that
∑n
i=1
xip/n = AM(xp 1 ,xp 2 ,...,xpn)
≤ rM(xp 1 ,xp 2 ,...,xpn)
=
Ñn
∑
i=1
(xpi)q/p/n
ép/q
=
Ñ n
∑
i=1
xqi/n
ép/q
.
Taking thep-th roots of both sides yields what we were after, viz.,
pM(x 1 ,x 2 ,...,xn) =
Ñ n
∑
i=1
xip/n
é 1 /p
≤
Ñn
∑
i=1
xqi/n
é 1 /q
= qM(x 1 ,x 2 ,...,xn).
Exercises.
- Show how Young’s inequality proves that GM(x 1 ,x 2 )≤AM(x 1 ,x 2 ),
wherex 1 ,x 2 ≥0. - Use Jensen’s inequality and the fact that the graph ofy= lnxis
concave down to obtain a simple proof that
AM(x 1 ,x 2 ,...,xn)≥GM(x 1 ,x 2 ,...,xn),
wherex 1 ,x 2 ,...,xn≥0.
- Use Jensen’s inequality to prove that given interior anglesA, B,
andC of a triangle then
sinA+ sinB+ sinC≤ 3
√
2 / 2.
Conclude that for a triangle 4 ABC inscribed in a circle of radius
R, the maximum perimeter occurs for an equilateral triangle. (See
Exercise 2 on page 34.)
- Given 4 ABCwith areaKand side lengthsa, b,andc, show that
ab+ac+bc≥ 4
√
3 K.
Under what circumstances does equality obtain? (Hint: note that
6 K =absinC+acsinB+bcsinA; use Cauchy-Schwarz together
with Exercise 3, above.)