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:

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

An equation!

June 21, 2024June 21, 2024 Tolaso

Let $f(x) = \left(x^2 +1 \right) \ln x \; , \; x>0$ .

Prove that $f$ is strictly increasing on $(0, +\infty)$ .
If $x \in \left(0, \frac{\pi}{2} \right)$ , solve the equation
$\cos^2 x \cdot f \left ( \tan x \right ) + \sin^2 x \cdot f \left ( \cot^{22} x \right ) =0$

Solution

It is $\displaystyle f'(x) = x \left ( \ln x^2 + 1 + \frac{1}{x^2} \right )$ . However, recalling that
$\ln x \geq 1 - \frac{1}{x} \quad \text{foreach} \quad x>0$

we conclude that $f$ is strictly increasing.
We note that
$\begin{align*} \cos^2 x \cdot f \left ( \tan x \right ) + \sin^2 x \cdot f \left ( \cot^{22} x \right ) =0 &\Leftrightarrow \frac{\cos^2 x}{\sin^2 x} \cdot f \left ( \tan x \right ) + f \left ( \cot^{22}x \right ) = 0 \\ &\Leftrightarrow \cot^2 x \cdot f \left ( \frac{1}{\cot x} \right ) + f \left ( \cot^{22}x \right ) =0 \\ &\!\!\!\!\!\!\!\!\!\overset{u = \cot x}{\Leftarrow \! =\! =\! =\! =\! \Rightarrow } u^2 f \left ( \frac{1}{u} \right ) + f \left ( u^2 \right ) = 0 \end{align*}$

On the other hand, $f \left ( \frac{1}{x} \right ) = - \frac{f(x)}{x^2}$ and hence:

$\begin{align*} f \left ( u^{22} \right ) = f(u) &\Leftrightarrow u^{22} = u \\ &\Leftrightarrow u = 1 \\ &\Leftrightarrow x = \frac{\pi}{4} \end{align*}$