# 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.