Inequality with Csc And Sin

Here's an inequality from the Problem Section of the American Mathematical Monthly [2014, 83]. The problem has been posted by Carol Kempiak, Aliso Niguel High School, Aliso Viejo, CA, and Bogdan Suceav˘a, California State University, Fullerton, CA. The solution (Solution 1) by Boris Karaivanov, Lexington, SC, was published in v 122, n 9, November 2015, p 905.

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 $z,$


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 $x=\csc A,$ $y=\csc B,$ $z=\csc C.$


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


What Is Red Herring

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