Minimum value of a fraction

Let $f(x) = \dfrac {13+12x}{\sqrt{1-x^2}}$ . For $f(x)$ to be real, $-1 < x < 1$ . We can let $x= \cos \theta$ , since $-1 \le \cos \theta \le 1$ . Then

$\begin{aligned} f(x) & = \frac {13+12\cos \theta}{\sin \theta} & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ & = \frac {13+\frac {12(1-t^2)}{1+t^2}}{\frac {2t}{1+t^2}} \\ & = \frac {13(1+t^2) + 12(1-t^2)}{2t} \\ \implies f(t) & = \frac {25 + t^2}{2t} = \frac {25}{2t} + \frac t2 \end{aligned}$

By AM-GM inequality :

$\begin{aligned} \frac {25}{2t} + \frac t2 & \ge 2 \sqrt{\frac {25}4} = 5 \\ \implies f(t) & \ge \boxed 5 \end{aligned}$

Equality occurs when $\dfrac {25}{2t} = \dfrac t2 \implies t = 5 \implies x = \cos \theta = \dfrac {1-t^2}{1+t^2} = - \dfrac {12}{13}$ .

Let the fraction be $A$ . In order that $A$ is definable, $1 - x^2 > 0 \Leftrightarrow x^2 < 1 \Leftrightarrow -1 < x < 1$ .

Working on the numerator of $A$ , we have:

$13+12x = (12,5 + 12,5x) + (0,5 - 0,5x) = 12,5(1+x) + 0,5(1-x) \geq 2\sqrt{(12,5)(0,5)(1-x)(x+1)} = 5\sqrt{1-x^2}$ (AM-GM inequality)

Therefore, we have:

$A=\frac{13+12x}{\sqrt{1-x^2}} \geq \frac{5\sqrt{1-x^2}}{\sqrt{1-x^2}} = 5$ (Note that $1-x^2 > 0$ )

Hence, the minimum value of $A$ is $\boxed{5}$ .

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The minimum value of $A$ is achieved only when $12,5 + 12,5x = 0,5 - 0,5x \Leftrightarrow 13x = -12 \Leftrightarrow x=-\frac{12}{13}$ (satisfies the condition)

Our expression can be written as (25(1+x)+(1-x))/2((1-x²)^1/2) =(1/2)*(25 ((1+x/1-x)^1/2) +((1-x/1+x)^1/2)) Taking ((1+x/1-x)^1/2)=t, we have (25t+1/t)/2 in which we can easily apply AM-GM inequality. Thus we get value of t as 1/5, which on solving for x gives x=-12/13