An Inequality in Triangle with the Sines of Half-Angles and Cube Roots

Problem

An Inequality in Triangle with the Sines of Half-Angles and Cube Roots

Solution 1

$\displaystyle\begin{align} 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}&=\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)2\sin^2 \frac{C}{2}\\ &=\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)(1-\cos C)\\ &=\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)-\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\cos C \end{align}$

(1)

$\displaystyle 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}=\sum_{cycl} \frac{a}{b}+\sum_{cycl} \frac{b}{a}-\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\cos C$

$\displaystyle\begin{align} \sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\cos C&=\sum_{cycl} \frac{a^2+b^2}{ab}\cdot \frac{a^2+b^2-c^2}{2ab}\\ &=\frac{1}{2a^2b^2c^2}\sum_{cycl} c^2(a^2+b^2)(a^2+b^2-c^2)\\ &=\frac{1}{2a^2b^2c^2}\sum_{cycl} \left[c^2(a^2+b^2)^2-c^4(a^2+b^2)\right]\\ &=\frac{1}{2a^2b^2c^2}\sum_{cycl} \left(c^2(a^4+b^4+2a^2b^2)-c^4a^2-c^4b^2\right)\\ &=\frac{1}{2a^2b^2c^2}\sum_{cycl} (c^2a^4+c^2b^4+2a^2b^2c^2-c^4b^2-c^4a^2)\\ &=\small{\frac{1}{2a^2b^2c^2}\left(\sum_{cycl} a^4c^2-\sum_{cycl} a^4c^2+\sum_{cycl} b^4c^2-\sum_{cycl} b^4c^2+6a^2b^2c^2\right)}\\ &=\frac{6a^2b^2c^2}{6a^2b^2c^2}=3 \end{align}$

We continue:

(2)

$\displaystyle 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}=\sum_{cycl} \frac{a}{b}+\sum_{cycl} \frac{b}{a}-3$


(3)

$\displaystyle \frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\sqrt[3]{\frac{a}{b}\cdot \frac{a}{b}\cdot \frac{b}{c}}=3\sqrt[3]{\frac{a^2}{bc}}=3\frac{a}{\sqrt[3]{abc}}$


(4)

$\displaystyle \frac{b}{c}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3]{\frac{b}{c}\cdot \frac{b}{c}\cdot \frac{c}{a}}=3\sqrt[3]{\frac{b^2}{ac}}=3\frac{b}{\sqrt[3]{abc}}$


(5)

$\displaystyle\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\geq 3\sqrt[3]{\frac{c}{a}\cdot \frac{c}{a}\cdot \frac{a}{b}}=3\sqrt[3]{\frac{c^2}{ab}}=3\frac{c}{\sqrt[3]{abc}}$


Further,

$\displaystyle\begin{align}&\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq 3\frac{a+b+c}{\sqrt[3]{abc}}\\ &\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b+c}{\sqrt[3]{abc}}\geq \frac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}=3 \end{align}$

From (2) it follows that

$\displaystyle\begin{align} 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}&=\sum_{cycl} \frac{a}{b}+\sum_{cycl} \frac{b}{a}-3\\ &\geq 3+\sum_{cycl} \frac{b}{a}-3\\ &=\sum_{cycl} \frac{b}{a}, \end{align}$

i.e.,

(6)

$\displaystyle 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}\ge \sum_{cycl} \frac{b}{a}.$

$\displaystyle \frac{a}{c}+\frac{a}{c}+\frac{b}{a}\geq 3\sqrt[3]{\frac{a}{c}\cdot \frac{a}{c}\cdot \frac{b}{a}}=\frac{3\sqrt[3]{abc}}{c}$

$\displaystyle \frac{b}{a}+\frac{b}{a}+\frac{c}{b}\geq 3\sqrt[3]{\frac{b}{a}\cdot \frac{b}{a}\cdot \frac{c}{b}}=3\frac{\sqrt[3]{abc}}{a}$

$\displaystyle \frac{c}{b}+\frac{c}{b}+\frac{a}{c}\geq 3\sqrt[3]{\frac{c}{b}\cdot \frac{c}{b}\cdot \frac{a}{c}}=3\frac{\sqrt[3]{abc}}{b}$

(7)

$\displaystyle \sum_{cycl} \frac{b}{a}\geq \sqrt[3]{abc}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

From (6) and (7),

$\displaystyle 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}\geq \sqrt[3]{abc}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

Solution 2

We'll prove instead a stronger inequality

$\displaystyle\sum_{cycl}\left(\frac{a}{b}+\frac{b}{a}\right)(1-\cos C)\ge\frac{a+b+c}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$

Or, equivalently,

$\displaystyle A-B\ge\frac{a+b+c}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right),$

where $A=\displaystyle\sum_{cycl}\left(\frac{a}{b}+\frac{b}{a}\right)\,$ and $B=\displaystyle\sum_{cycl}\left(\frac{a}{b}+\frac{b}{a}\right)\cos C.$

$\displaystyle\begin{align} B &= \sum_{cycl}\left(\frac{a}{b}+\frac{b}{a}\right)\cos C\\ &=\sum_{cycl}\left(\frac{a^2+b^2}{ab}\right)\left(\frac{a^2+b^2-c^2}{2ab}\right)\\ &=\sum_{cycl}\left(\frac{(a^2+b^2)^2}{2a^2b^2}\right)-\left(\frac{c^2(a^2+b^2)}{2a^2b^2}\right)\\ &=\sum_{cycl}\frac{a^2}{2b^2}+\sum_{cycl}\frac{b^2}{2a^2}+\sum_{cycl}1-\sum_{cycl}\frac{c^2}{2b^2}-\sum_{cycl}\frac{c^2}{2a^2}\\ &=3. \end{align}$

$A=\displaystyle\sum_{cycl}\left(\frac{a}{b}+\frac{b}{a}\right)=\sum_{cycl}\frac{b+c}{a}.$

We need to show that

$\displaystyle A-B=\sum_{cycl}\frac{b+c}{a}-3\ge\left(\frac{a+b+c}{3}\right) \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$

This is equivalent to

$\displaystyle \left(\sum_{cycl}a\right)\left(\sum_{cycl}\frac{1}{a}\right)-6\ge\left(\frac{a+b+c}{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right),$

i.e.,

$\displaystyle \frac{2}{3}\left(\sum_{cycl}a\right)\left(\frac{1}{a}\right)\ge 6,$

which is true because, by the AM-GM inequality,

$\displaystyle \left(\sum_{cycl}a\right)\left(\frac{1}{a}\right)\ge 3\sqrt[3]{abc}\cdot 3\frac{1}{\sqrt[3]{abc}}=9.$

Solution 3

First observe that

$\displaystyle\begin{align} LHS&= 2\sum_{cycl} \left(\frac{a}{b}+\frac{b}{a}\right)\sin^2 \frac{C}{2}\\ &= 2\sum_{cycl} \left(\frac{a^2+b^2}{ab}\right)\frac{(s-a)(s-b)}{ab}\\ &= \sum_{cycl} \frac{c^2(a^2+b^2)(b+c-a)(c+a-b)}{2a^2b^2}. \end{align}$

Let's prove that

(1)

$\displaystyle\begin{align}LHS&=\sum_{cycl} \frac{c^2(a^2+b^2)(b+c-a)(c+a-b)}{2a^2b^2}\\ &\ge\left(\frac{a+b+c}{3}\right)\left(\frac{ab+bc+ca}{abc}\right). \end{align}$

This is equivalent to

$\displaystyle\begin{align} &3\sum_{cycl}c^2(a^2+b^2)(b+c-a)(c+a-b)\ge 2abc\sum_{cycl}a\sum_{cycl}ab\,\Longleftrightarrow\\ &4(a^3b^2c+a^3bc^2+b^3c^2a+b^3ca^2+c^3a^2b+c^3ab^2)\ge 24a^2b^2c^2\,\Longleftrightarrow\\ &(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)\ge 6abc, \end{align}$

which is the same as

(2)

$\displaystyle b(a^2+c^2)+c(a^2+b^2)+a(b^2+c^2)\ge 6abc.$

But, by the AM-GM inequality, $\displaystyle b(a^2+c^2)\ge 2abc,\,$ $\displaystyle c(a^2+b^2)\ge 2abc,\,$ $\displaystyle a(b^2+c^2)\ge 2abc,\,$ so that (2) holds and so is (1).

This is stronger than the required inequality.

Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem of his that was published in the Romanian Mathematical Magazine. Dan messaged me his solution (Solution 1) in a tex file. Solution 2 is by Kevin Soto Palacios; Solution 3 is by Soumava Chakraborty.

 

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