# Inequality with Csc And Sin

Here's an inequality from the Problem Section of the *American Mathematical Monthly*

In a triangle with angles of radian measure $A,$ $B,$ $C,$ prove that

$\displaystyle\frac{\csc A+\csc B+\csc C}{2}\ge\\ \;\;\;\displaystyle\frac{1}{\sin B+\sin C}+\frac{1}{\sin C+\sin A}+\frac{1}{\sin A+\sin B},$

with equality if and only if the triangle is equilateral.

### Solution 1

Use the arithmetic mean-harmonic mean inequality (see examples). For any positive $x,$ $y,$ and $z,$

$\displaystyle\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{x}+\frac{1}{y}}{2}+\frac{\frac{1}{y}+\frac{1}{z}}{2}+\frac{\frac{1}{z}+\frac{1}{x}}{2}\ge\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x},$

with equality only if $x=y=z.$ Set $x=\sin A,$ $y=\sin B,$ $z=\sin C.$

### Solution 2

$\displaystyle x+y+z\ge\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}\ge\frac{2}{\frac{1}{x}+\frac{1}{y}}+\frac{2}{\frac{1}{y}+\frac{1}{z}}+\frac{2}{\frac{1}{z}+\frac{1}{x}},$

with equality only if $x=y=z.$ Set $x=\csc A,$ $y=\csc B,$ $z=\csc C.$

### Remark

Clearly the stipulation that $A,$ $B,$ $C$ are the angle measures in a triangle is a red herring. The inequality will hold for, say, $A=1,$ $B=2,$ $C=3.$ To disguise the simple solution the inequality cold have been presented, e.g., as

$\displaystyle\frac{x+y+z}{2}\ge\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x}.$

### What Is Red Herring

- On the Difference of Areas
- Area of the Union of Two Squares
- Circle through the Incenter
- Circle through the Incenter And Antiparallels
- Circle through the Circumcenter
- Inequality with Logarithms
- Breaking Chocolate Bars
- Circles through the Orthocenter
- 100 Grasshoppers on a Triangular Board
- Simultaneous Diameters in Concurrent Circles
- An Inequality from the 2015 Romanian TST
- Schur's Inequality
- Further Properties of Peculiar Circles
- Inequality with Csc And Sin
- Area Inequality in Trapezoid
- Triangles on HO
- From Angle Bisector to 120 degrees Angle
- A Case of Divergence
- An Inequality for the Cevians through Spieker Point via Brocard Angle
- An Inequality In Triangle and Without
- Problem 3 from the EGMO2017
- Mickey Might Be a Red Herring in the Mickey Mouse Theorem
- A Cyclic Inequality from the 6th IMO, 1964
- Three Complex Numbers Satisfy Fermat's Identity For Prime Powers
- Probability of Random Lines Crossing
- Planting Trees in a Row
- Two Colors - Three Points

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny67608587