What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq  3(  \sec^6 x \csc^6 x \sec^6 x\csc^6 x)^\frac{1}{3}   $$
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x \geq 3(  \sec x \csc)^4 $$
$$
\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq
 \frac{3 * 2^4}{\sin ^4 2x} $$
Clearly least value is 48, but something is wrong here, as the answer is 80, if I use other methods.
 A: You want to find the least value of $f(x)=\sec^6(x)+\csc^6(x)+\sec^6(x)\csc^6(x)$. You found that $f(x) \geq g(x)=3(\sec(x)\csc(x))^4$. In addition, the minimum value of $g(x)$ is $48$. Therefore, you can conclude that $f(x) \geq 48$ for all $x$. But why would you expect there to exist some $x$ such that $f(x)=48$, when $g(x)$ was simply a lower bound?
This is like saying find the least value of $x^2+4$. Well, $x^2+4 \geq (4x^2)^{1/2} = 4|x|$, whose minimum value is $0$. But clearly $x^2+4$ has a minimum value of $4$. The problem is that the lower bound is not tight.
A: Others explain why $48$ is correct as a a lower bound but may not be the sharp lower bound.
One way to get a lower bound of $80$ involves using the fact that each term is a cubed quantity.  Start with the decomposition
$\sec^6 x + \csc^6 x + \sec^6 x\csc^6 x=A+B$
$A=\sec^6 x + \csc^6 x$
$B=\sec^6 x\csc^6 x$
Factor $A$ as a sum of cubes:
$A=(\sec^2 x + \csc^2 x)(\sec^4 x - \sec^2 x\csc^2 x + \csc^4 x)$
$A=\dfrac{(\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x\sin^2 x + \sin^4 x)}{\cos^6 x\sin^6 x}$
Plugging in $\cos^2 x +\sin^2 x =1$ and $\cos^4 x +2 \cos^2 x\sin^2 x + \sin^4 x=(\cos^2 x +\sin^2 x)^2=1$:
$A=\dfrac{1 - 3\cos^2 x\sin^2 x}{\cos^6 x\sin^6 x}$
We have $(\cos x-\sin x)^2=1-2\cos x\sin x\ge 0$ forcing $|\cos x\sin x|\le 1/2$.  Thereby
$A\ge (1 - 3/4)×(64)=16$
For $B$, simply render
$B=\dfrac{1}{\cos^6 x\sin^6 x}\ge 64$
where again we have put in $|\cos x\sin x|\le 1/2$.
Then
$A+B\ge 16+64=80$.
This bound may be proven sharp by putting in $x=\pi/4$, or by noting that the separate bounds on $A$ and $B$ both become sharp when $|\cos x|=|\sin x|$.
A: In your way you proved that the minimal value is greater than $48$. 
It's true, but the equality does not occur, which says that $48$ is not a minimal value.
The right solution can be the following, for example.
Let $\sin^2x\cos^2x=t.$
Thus, by AM-GM
$$t\leq\left(\frac{\sin^2x+\cos^2x}{2}\right)^2=\frac{1}{4}.$$
The equality occurs for $x=45^{\circ},$ which gives a value $80$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\frac{\sin^4x-\sin^2x\cos^2x+\cos^4x}{\sin^6x\cos^6x}+\frac{1}{\sin^6x\cos^6x}\geq80$$ or
$$\frac{1-3\sin^2x\cos^2x}{\sin^6x\cos^6x}+\frac{1}{\sin^6x\cos^6x}\geq80$$ or
$$\frac{2}{t^3}-\frac{3}{t^2}\geq80$$ or
$$80t^3+3t-2\leq0$$ or
$$80t^3-20t^2+20t^2-5t+8t-2\leq0$$ or
$$(4t-1)(20t^2+5t+2)\leq0,$$ which is obvious.
A: Alternatively:
$$\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x=\frac{\sin^6x+\cos^6x+1}{\sin^6 x\cos^6x}=\\
\frac{2^6[(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)+1]}{\sin^62x}=\\
\frac{64(2-\frac34\sin^22x)}{(\sin^22x)^3}=\frac{128-48(1-\cos^22x)}{(\sin^22x)^3}=\\
\frac{80+48\cos^22x}{(\sin^22x)^3}\ge 80.$$
equality occurs for $\sin2x=\pm1$.
