Lorian Saceanu's Sides And Angles Inequality

Problem

Lorian Saceanu's Sides And Angles Inequality, problem

Solution 1

The left inequality is well known and is a one-step application of Chebyshev's inequality.

For the right-hand side, observe that

$\begin{align} a&=b\cos\gamma+c\cos\beta\\ b&=c\cos\alpha+a\cos\gamma\\ c&=a\cos\beta+b\cos\alpha. \end{align}$

Summing up,

$(a+b+c)(\cos\alpha+\cos\beta+\cos\gamma-1)=a\cos\alpha+b\cos\beta+c\cos\gamma.$

Now, it is known that $\displaystyle \cos\alpha+\cos\beta+\cos\gamma=1+\frac{r}{R}.$ Combining the latter two identities,

$\displaystyle \frac{a}{a+b+c}\cos\alpha+\frac{b}{a+b+c}\cos\beta+\frac{c}{a+b+c}\cos\gamma=\frac{r}{R}.$

The weighted Jensen's inequality, yields

$\displaystyle \frac{a}{a+b+c}\cos\alpha+\frac{b}{a+b+c}\cos\beta+\frac{c}{a+b+c}\cos\gamma\le \cos\frac{a\alpha+b\beta+c\gamma}{a+b+c}$

so that $\displaystyle \frac{r}{R}\le\cos\frac{a\alpha+b\beta+c\gamma}{a+b+c}$ from which, applying $\arccos$ to both sides we obtain the required

$\displaystyle \frac{a\alpha + b\beta + c\gamma}{a+b+c}\le \arccos\left(\frac{r}{R}\right).$

Solution 2

The triangle is acute angled. Hence, the sequences $\{\sin \alpha, \sin \beta, \sin\gamma \}$ and $\{\alpha, \beta, \gamma\}$ are similarly sorted. By sine rule, $\{a, b, c\}$ also has the same sorting. Thus, by the rearrangement inequality,

$\displaystyle \begin{align} a\alpha+b\beta+c\gamma&\geq a\left(\frac{\beta + \gamma}{2}\right) + b\left(\frac{\gamma + \alpha}{2}\right) + c\left(\frac{\alpha + \beta}{2}\right) \\ &\geq a\left(\frac{\pi-\alpha}{2}\right) + b\left(\frac{\pi-\beta}{2}\right) + c\left(\frac{\pi-\gamma}{2}\right). \\ \frac{a\alpha+b\beta+c\gamma}{a+b+c}&\geq \frac{\pi}{3}.~\text{(Rearranging the terms)} \end{align}$

Using Jensen's inequality

$\displaystyle \begin{align} \cos\left(\frac{a\alpha+b\beta+c\gamma}{a+b+c}\right) &\geq \frac{a\cos\alpha+b\cos\beta+c\cos\gamma}{a+b+c} \\ &= \frac{\sin\alpha\cos\alpha+\sin\beta\cos\beta+\sin\gamma\cos\gamma}{ \sin\alpha+\sin\beta+\sin\gamma}~\text{(sine rule)} \\ &= \frac{\sin 2\alpha+\sin 2\beta+\sin 2\gamma}{2(\sin\alpha+\sin\beta+\sin\gamma)} \\ &= \frac{\sin\alpha\sin\beta\sin\gamma}{2\cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) \cos\left(\frac{\gamma}{2}\right)}~\text{(Using identities for angles in a triangle)} \\ &= 4\sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) \sin\left(\frac{\gamma}{2}\right)=\frac{r}{R}. \end{align}$

Cosine being a monotonically decreasing function for acute angles,

$\displaystyle \frac{a\alpha+b\beta+c\gamma}{a+b+c}\leq \arccos\left(\frac{r}{R}\right).$

Acknowledgment

This problem by Lorian Saceanu was privately communicated to me by its author, along with a solution of his (Solution 1.) Solution 2 is by Amit Itagi.

 

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