How to Prove the Trigonometric Inequality for Real Numbers?

[anemone]

For real numbers $$0\lt x\lt \frac{\pi}{2}$$, prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$.

 

[MarkFL]

My solution:

Spoiler

If we observe that:

$$g(x)=\sin^2(x)\tan(x)$$

$$h(x)=\cos^2(x)\cot(x)$$

are complimentary functions, we can state the problem as:

Optimize the objective function:

$$f(x,y)=\sin^2(x)\tan(x)+\sin^2(y)\tan(y)$$

Subject to the constraints:

$$x+y=\frac{\pi}{2}$$

$$0<x,\,y<\frac{\pi}{2}$$

And so...wait for it...can you guess where I'm going with this?...Yes! By cyclic symmetry, we know the critical point is:

$$(x,y)=\left(\frac{\pi}{4},\frac{\pi}{4}\right)$$

And we then find:

$$f\left(\frac{\pi}{4},\frac{\pi}{4}\right)=1$$

Checking another point on the constraint, such as:

$$(x,y)=\left(\frac{\pi}{6},\frac{\pi}{3}\right)$$

We find:

$$f\left(\frac{\pi}{6},\frac{\pi}{3}\right)=\frac{5}{2\sqrt{3}}>1$$

And so we know:

$$f_{\min}=1$$

 

[Greg]

For real numbers $$0\lt x\lt \frac{\pi}{2}$$, prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$.

Spoiler

$$\cos^2(x)\cot(x)+\sin^2(x)\tan(x)=(1-\sin^2(x))\cot(x)+(1-\cos^2(x))\tan(x)$$

$$=\cot(x)+\tan(x)-\sin(2x)=\dfrac{2-\sin^2(2x)}{\sin(2x)}=\dfrac{1+\cos^2(2x)}{\sin(2x)}\quad(1)$$

As $0<\sin(2x)\le1$ and $1\le1+\cos^2(2x)<2$, $(1)\ge1$ with equality at $x=\dfrac{\pi}{4}$.

 

[anemone]

Hi MarkFL and greg1313! Thanks for participating and good job to the both of you! (Cool)

My solution:

Spoiler

First, note that we can rewrite $\cos^2 x \cot x+\sin^2 x \tan x$ as $$\frac{\cos^3 x}{\sin x}+\frac{\sin^3 x}{\cos x}$$.

For the domain $$0\lt x\lt \frac{\pi}{4}$$, we have $\cos^3 x\gt \sin^3x,\,\dfrac{1}{\sin x}\gt \dfrac{1}{\cos x}$, so by the rearrangement inequality we have:

$\begin{align*}\cos^2 x \cot x+\sin^2 x \tan x=\dfrac{\cos^3 x}{\sin x}+\dfrac{\sin^3 x}{\cos x}&\ge \dfrac{\cos^3 x}{\cos x}+\dfrac{\sin^3 x}{\sin x}\\&=\cos^2x+\sin^2 x\\&=1\end{align*}$

By the same token, for the domain $$\frac{\pi}{4}\le x\lt \frac{\pi}{2}$$, we have $\sin^3 x\gt \cos^3x,\,\dfrac{1}{\cos x}\gt \dfrac{1}{\sin x}$, so by the rearrangement inequality we also have:

$\begin{align*}\cos^2 x \cot x+\sin^2 x \tan x=\dfrac{\sin^3 x}{\cos x}+\dfrac{\cos^3 x}{\sin x}&\ge \dfrac{\sin^3 x}{\sin x}+\dfrac{\cos^3 x}{\cos x}\\&=\sin^2x+\cos^2 x\\&=1\end{align*}$

Combining the two yields the result.

 

[lfdahl]

My suggested solution:

Spoiler

By the AM-GM inequality, we get:

$\frac{\cos^3x}{\sin x}+\frac{sin^3x}{\cos x} \ge 2\sqrt{\frac{\cos^3x\sin^3x}{\cos x\sin x}} = \sin 2x \le 1, \;\;\; 0 < x < \frac{\pi}{2}$

Thus

$\cos^2x\cot x+\sin^2x\tan x \ge 1$

 

[kaliprasad]

My suggested solution:

Spoiler

By the AM-GM inequality, we get:

$\frac{\cos^3x}{\sin x}+\frac{sin^3x}{\cos x} \ge 2\sqrt{\frac{\cos^3x\sin^3x}{\cos x\sin x}} = \sin 2x \le 1, \;\;\; 0 < x < \frac{\pi}{2}$

Thus

$\cos^2x\cot x+\sin^2x\tan x \ge 1$

I aam sorry if I misunderstood but

Spoiler

$a > b $ and $ b <=c$ does not mean $ a > c$

 

[lfdahl]

I aam sorry if I misunderstood but

Spoiler

$a > b $ and $ b <=c$ does not mean $ a > c$

Spoiler

You´re right, kaliprasad. The question is, what can I conclude from:

$LHS \ge \sin2x$?

Anyway, the LHS has its minimum (by symmetry) at $x = \frac{\pi}{4}$ $(\cos x = \sin x = \frac{1}{\sqrt{2}})$.
i.e. $LHS_{min} = 1$.