How prove this $\tan{x}>\frac{3x}{2+\sqrt{1-x^2}}$

Question

let $0<x<1$,prove that $$\tan{x}>\dfrac{3x}{2+\sqrt{1-x^2}}$$

This problem have nice solution?

my idea: let $$f(x)=\tan{x}-\dfrac{3x}{2+\sqrt{1-x^2}}=\tan{x}-3x\dfrac{2-\sqrt{1-x^2}}{3+x^2}$$

and other idea: let $x=\cos{t}$,then $$\tan{(\cos{x})}>\dfrac{3\cos{t}}{2+\sin{t}}$$

other (2) idea: $$\tan{x}>x+\dfrac{1}{3}x^3$$

Your (2) idea does not work close to $x=1$, as $\displaystyle x+\frac{x^3}{3} \approx \frac{4}{3} < \frac{3x}{2+\sqrt{1-x^2}}\approx\frac{3}{2}$ near $x=1$. — Macavity, Aug 18 '13 at 03:56
If I were to attack this, I would try to show that $\dfrac{x\cos x}{\sin x}\lt \dfrac{2+\sqrt{1-x^2}}{3}$. — André Nicolas, Aug 18 '13 at 03:57
If we had tried to insist on the taylor expansion of $\tan(x)$, we would have to use at least four terms — chubakueno, Aug 18 '13 at 04:38

André Nicolas · Accepted Answer · 2013-08-18T04:55:53.127

8

\We show equivalently that if $0\lt x\lt 1$, then $$x\cot x\lt \frac{2+\sqrt{1-x^2}}{3}.\tag{1}$$ We use the fact that $x\cot x\lt 1-\frac{x^2}{3}$ in our interval. So we want to show that $$1-\frac{x^2}{3}\lt \frac{2+\sqrt{1-x^2}}{3}.\tag{2}$$

Multiply through by $3$, and rearrange. Inequality (2) is equivalent to $1-x^2\lt \sqrt{1-x^2}$, which is clear.

edited Aug 18 '13 at 04:55

answered Aug 18 '13 at 04:24

André Nicolas

507,029

Does $x\cot x\lt 1-\frac{x^2}{3}$ come from Taylor series? – Ovi Aug 18 '13 at 04:30
@Ovi (http://www.wolframalpha.com/input/?i=taylor+expansion+of+cotx) – chubakueno Aug 18 '13 at 04:35
Yes, it is Taylor series. – André Nicolas Aug 18 '13 at 04:38
@AndréNicolas Ok thanks – Ovi Aug 18 '13 at 17:44

score 1 · Answer 2 · answered Aug 18 '13 at 05:12

Your idea works equally well too.

We need to prove that $$ \cos{t} + \frac{\cos{t}^3}{3} > \frac{3\cos{t}}{2+\sin{t}} $$ Since $x$ varies between $(0,1)$, both sin(t) and cos(t) vary from (0,1). That reduces our inequality to: $$ 1 + \frac{\cos^2{t}}{3} > \frac{3}{2+\sin{t}} $$

Expressing cos in terms of sin, and further simplifying we get: $$ (2+\sin{t})^2(2-\sin{t}) > 3 $$

We substitute $y = \sin{t}$, where $y \in (0,1)$: $$ -y^3 -2y^2 + 4y + 8 > 3 $$

Analyzing the derivative of the LHS, we can easily find that the minimum value of the polynomial in $(0,1)$ is $8$, and therefore the inequality is true in the given interval.

should be $ (2+\sin{t})^2(2-\sin{t}) > 9 $ (9 instead of 3) and this in not true... — larry01, Jul 19 '16 at 20:05

score 0 · Answer 3 · answered Aug 18 '13 at 05:43

why dont you go for the calculus approach.

assume $ f(x) = tanx-3x/2+\sqrt{1+x^2} $.

differentiate w.r.t to x

we get $ f'(x) = sec^{2}x - 3/(2+\sqrt{1+x^2} + 3(x^2)/(2+\sqrt{1+x^2})^2 * 1/\sqrt{1+x^2} $ ,

under the given interval , $ f'(x) > 0 $ , implying it is an increasing function. we know that $f'(x) > 0 <=> f(x) > 0 .$.

thus $ tanx > 3x/1+\sqrt{+x^2} $

How prove this $\tan{x}>\frac{3x}{2+\sqrt{1-x^2}}$

3 Answers3