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.

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 $|\cos x|=|\sin x|$.
