Page 44 - RMGO 5

44 Mih´aly BENCZE

Theorem 2. In any convex polygon A 1 A 2 . . . A n hold the inequalities:

X a 1 n
1. ≥
a 2 + a 3 + . . . + a n n − 1
(Nesbitt’s Inequality);

a 1
X n
2. ≥
−a 1 + a 2 + . . . + a n n − 2
(another generalization of problem M 7, Kvant);
n
X
a k
X a 2 1 k=1
3. ≥ ;
a 2 + a 3 + . . . + a n n − 1
n
X
a k
X a 2 k=1
4. 1 ≥ .
−a 1 + a 2 + . . . + a n n − 2
n
X
Proof. Let S = a k .
k=1
x 2S
00
1) The function f(x) = is convex because f (x) = > 0. Using
S − x (S − x) 3
the Jensen’s Inequality holds the result.
x 4S
00
2) The function g(x) = is convex because g (x) = > 0.
S − 2x (S − 2x) 3
Using the Jensen’s Inequality holds the result.
3) By Cauchy-Schwarz Inequality we get

n n n
! 2 ! 2
X X X
a k a k a k
X a 2 1 k=1 k=1 k=1
≥ X = n = .
a 2 + a 3 + . . . + a n (a 2 + a 3 + . . . + a n ) X n − 1
(n − 1) a k
k=1

4) By Cauchy-Schwarz Inequality we get

n n n
! 2 ! 2
X X X
a k a k a k
X a 2
1 k=1 k=1 k=1
≥ X = n = .
−a 1 + a 2 + . . . + a n (−a 1 + a 2 + . . . + a n ) X n − 2
(n − 2) a k
k=1