Joy of Mathematics

An inequality

December 14, 2024 Tolaso

Let $a,b,c \in (0, 1)$ such that $ab + bc + ca =1$ . Prove that

$\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$

Solution

We have successively:

$\begin{align*} \sum_{cyc} \frac{a^2+b^2}{(1-a)^2(1-b^2)} &\geq \sum_{cyc}\frac{\frac{1}{2}(a+b)^2}{(1-ab)^2} \\ & = \sum_{cyc}\frac{\frac{1}{2}(a+b)^2}{c^2(a+b)^2} \\ & = \frac{1}{2}\sum_{cyc}\frac{1}{a^2} \\ & \geq \frac{1}{2}\sum_{cyc}\frac{1}{ab} \\ & =\frac{1}{2}\sum_{cyc}ab\sum_{cyc}\frac{1}{ab} \\ & \geq\frac{9}{2}. \end{align*}$

An equality

December 14, 2024 Tolaso

Prove that, in any given triangle, it holds:

$(a+b+c)\left(\tan\frac{A}{2}+\tan\frac{B}{2}\right) = 2 c \cot\frac{C}{2}$

Solution

Using the fact that $\displaystyle \tan \frac{A}{2} = \sqrt{\frac{\left( s-b \right)\left( s-c \right)}{s \left( s-a \right)}}$ and similarly for the other number, where $s$ is the semiperimeter of the triangle we have successively:

$\begin{align*} \left( a + b +c \right) \left( \tan \frac{A}{2} + \tan \frac{B}{2} \right) & = 2s \left( \sqrt{\frac{\left( s-b \right) \left( s-c \right)}{s \left( s - a \right)}} + \sqrt{\frac{\left( s-c \right) \left( s-a \right)}{s \left( s - b \right)}} \right) \\ & = 2s \frac{\sqrt{s-c}}{\sqrt{s}} \left( \frac{2s - a-b}{\sqrt{s-a} \sqrt{s-b}} \right) \\ & = 2c \frac{\sqrt{s} \sqrt{s-c}}{\sqrt{\left( s -a \right) \left( s-b \right)}} \\ & = 2c \sqrt{\frac{s \left( s-c \right)}{\left( s-a \right) \left( s-b \right)}} \\ & = 2c \cot \frac{C}{2} \end{align*}$

A logarithm

August 18, 2023 Tolaso

Let $a \in (0, 1)$ . If $\log_a \left ( 1 - a \right ) = 5$ , prove that $\log_a \left ( 1 + a \right ) = -2$ .

An identity

July 23, 2023 Tolaso

If $\alpha + \beta + \gamma =0$ , prove that

$2 \left ( \alpha^5 + \beta^5 + \gamma^5 \right ) = 5 \alpha \beta \gamma \left ( \alpha^2 + \beta^2 + \gamma^2 \right )$

Trigonometry with golden ratio

March 4, 2023 Tolaso

Let $\varphi$ denote the golden ratio. Prove that

$4 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{\varphi^6 - 1}} = \frac{\pi}{2}$

Solution

Since $\varphi^2 = \varphi + 1$ we deduce that $\varphi^3 = \varphi^2 + \varphi = 2 \varphi + 1$ . Furthermore,

$\varphi^6 - 1 = \left ( \varphi^3 \right )^2 -1 = \left ( 2 \varphi + 1 \right )^2 -1 = 4\varphi^2 + 4 \varphi = 4 \varphi^3$

Also, taking into consideration that $\displaystyle 2 \arctan x = \arctan \frac{2x}{1-x^2} \; ,\; x^2<1$ we deduce that:

$\begin{align*} 4 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{\varphi^6 - 1}} &= 2 \cdot 2 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{4 \varphi^3}} \\ &= 2 \arctan \sqrt{\varphi} - \arctan \frac{1}{2 \sqrt{\varphi^3}} \\ &= \arctan \sqrt{\varphi} + \arctan \sqrt{\varphi} - \arctan \frac{1}{2 \sqrt{\varphi^3}} \\ &= \arctan \sqrt{\varphi} + \arctan \frac{1}{\sqrt{\varphi}} \\ &= \frac{\pi}{2} \end{align*}$

In the last step we made use of the identity $\displaystyle \arctan \alpha + \arctan \beta = \arctan \frac{x+y}{1-xy} \;,\; xy<1$ .