Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
 A: using the lemma $$t+\frac{1}{t}\geq 2$$ for $t>0$  we get the inequality $$\tan(x)+\frac{1}{\tan(x)}\geq 2$$
your way is also ok.
A: We already have that
$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$$
If $x$ is an acute angle strictly between $0$ and $\pi/2$, then $\sin x \cos x > 0$. Hence $\tan x + \cot x \geq 2$ if and only if $$1 \geq 2\sin x \cos x \ \ \ \ \  \text{  or alternatively }  \ \ \ \ \ \  \sin(2x) \leq 1$$
and this last result is true because $\sin\theta \leq 1$ for all angles $\theta$.

In other words, you almost had it. In your second line, you just had to realize that $$2\sin x \cos x$$ could be written as $\sin 2x$ and then bounded above by $1$.
A: Nice Work
Just don't start by assuming the equality or inequality


Indeed there is another way, There's an elementary way

Since we are only concerned about acute angles
$$\left(a-\frac1a\right)^2\ge0$$
$$a^2-2+\frac1{a^2}\ge0$$
$$a^2+\frac1{a^2}\ge2$$
put $a^2=\tan x$
$$\tan x+\frac1{\tan x}\ge2$$
$$\tan x+\cot x\ge2$$
$\text{Game Over!}$
A: Since $x$ is an acute angle, $\cot x$ and $\tan x$ are positive.
Therefore, by A.M.-G.M. inequality,
$\frac {\tan x + \cot x}{2} \ge \sqrt{\tan x \times \cot x}$
Now, $\tan x \times \cot x = 1$ (by definition)
Therefore we obtain,
$\tan x + \cot x \ge 2 \sqrt{1}$
And so, $\tan x + \cot x \ge 2 $
Proof complete.
A: It is quite easy...
$\tan x + \cot x$
= $\tan x + \cot x - 2 \sqrt {\tan x \cot x} + 2 \sqrt{\tan x \cot x}$ 
=$(\sqrt {\tan x} -\sqrt {\cot x})^2+ 2$   [because $\tan x \cot x =1$]

Since square of any number is greater than or equal to zero, we can write, 
$(\sqrt {\tan x} -\sqrt {\cot x})^2 \ge 0$ 
So, $(\sqrt {\tan x} -\sqrt {\cot x})^2 + 2 \ge 2$
But, $(\sqrt {\tan x} -\sqrt {\cot x})^2 + 2 = \tan x + \cot x$ 
So, $\tan x + \cot x \ge 2$
NOTE : We want any square number to make this problem easier, this is a standard technique for proving most of inequalities in mathematics, therefore we have added and subtract $2 \sqrt{\tan x \cot x}$  in the main expression to make the expression perfect square.
A: Let $ y = \tan x + \cot x $.
Then $ y' = \sec^{2} x - \csc^{2} x $.
So when $ y' = 0 $, $ x = \pi/4 $, for $x$ acute.
You can prove this is a minimum by considering $ y' $ for $ x = \pi/6 $ and $ x = \pi/3 $.
Thus $ y_{min} = \tan \pi/4 + \cot \pi/4 = 1 + 1 = 2.$
A: Even if your implications are correct, your logic is flawed.
Logically, what you have done is similar to
$$\begin{align}
+1&=-1
\\ \Rightarrow (+1)^2&=(-1)^2
\\ \Rightarrow 1&=1.
\end{align}$$
Now $1=1$ so $+1=-1$ must be true also.
Deriving a true statement says nothing about the truth of the hypothesis.
A: Notice that $\tan x\times \cot\ x=1$, and
according to the inequality of means we get $$\tan x+\cot x \geqslant 2\sqrt{\tan x\times \cot x}=2$$
which is what we wanted.
