The sum of the triangle sides lengths reciprocals vs a cyclic sum of a specific form.
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The document presents a mathematical proof demonstrating that for sides of a triangle a, b, and c, an inequality involving their reciprocals holds. It utilizes fundamental properties of triangle sides and algebraic manipulation to establish the result. The conclusion affirms the validity of the proposed inequality related to the sides of the triangle.
The sum of the triangle sides lengths reciprocals vs a cyclic sum of a specific form.
- 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