<s>[INST] 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.
 [/INST] 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.
</s>