Proof of the inequality between the sum of the reciprocals of a triangle sides lengths and a cyclic sum of a specific form. Use of a transformed inequality between the arithmetic mean and the harmonic mean.

1. Prove that if a, b, c are lengths of the sides of a triangle then the following inequality holds:
1
a
+
1
b
+
1
c
1
a + b − c
+
1
c + a − b
+
1
b + c − a
Solution by Mikołaj Hajduk: Since a, b, c are lengths of the sides of a triangle we have a > 0, b > 0, c > 0
and 


a + b > c
c + a > b
b + c > a
⇐⇒



a + b − c > 0
c + a − b > 0
b + c − a > 0
Let’s notice that
(a + b − c) + (c + a − b) = 2a
(c + a − b) + (b + c − a) = 2c
(a + b − c) + (b + c − a) = 2b
From the other hand, for any x > 0, y > 0 we have
1
x
+
1
y
4
x + y
hence
1
a + b − c
+
1
c + a − b
+
1
b + c − a
4
(a + b − c) + (c + a − b)
+
1
b + c − a
=
2
a
+
1
b + c − a
1
a + b − c
+
1
c + a − b
+
1
b + c − a
1
a + b − c
+
4
(c + a − b) + (b + c − a)
=
1
a + b − c
+
2
c
1
a + b − c
+
1
c + a − b
+
1
b + c − a
4
(a + b − c) + (b + c − a)
+
1
c + a − b
=
2
b
+
1
c + a − b
c 2015/10/11 22:05:55, Mikołaj Hajduk 1 / 2 next

2. Let’s add all aforementioned inequalities:
3
1
a + b − c
+
1
c + a − b
+
1
b + c − a
2
a
+
2
b
+
2
c
+
1
a + b − c
+
1
c + a − b
+
1
b + c − a
and rearrange the result a bit to obtain the following inequality:
2
1
a + b − c
+
1
c + a − b
+
1
b + c − a
2
1
a
+
1
b
+
1
c
and finally
1
a + b − c
+
1
c + a − b
+
1
b + c − a
1
a
+
1
b
+
1
c
c 2015/10/11 22:05:55, Mikołaj Hajduk 2 / 2