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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.

user541396
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    Put very simply: you defined a lower bound. But AM-GM alone does not guarantee the sharp lower bound. Understand this, then you can read the answers for how to identify the sharp bound. – Oscar Lanzi Mar 30 '19 at 15:50
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    https://math.stackexchange.com/questions/3166878/find-the-least-value-of-sec6-x-csc6-x-sec6-x-csc6-x – lab bhattacharjee Mar 30 '19 at 16:11
  • Search more precisely. I said "lower bound". Try to Google that, maybe also put "mathematics" into the search window. – Oscar Lanzi Mar 30 '19 at 16:39

4 Answers4

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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.

kccu
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  • What is wrong with $x^2$ not being positive? AM-GM works for all non-negative numbers. Edit: I had a typo with my inequality sign going the wrong way. Perhaps that was the cause of your confusion? – kccu Apr 04 '19 at 12:51
  • The very first sentence of the Wikipedia article on AM-GM is "In mathematics, the inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list" – kccu Apr 04 '19 at 14:25
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    As the Wikipedia article states, equality holds in AM-GM if and only if all the numbers are the same. Thus if you are applying AM-GM to functions, the inequality given by AM-GM is sharp if and only if there is a value of $x$ such that all the terms (in your case $\sec^6(x)$, $\csc^6(x)$, $\sec^6(x)\csc^6(x)$) have the same value (this is not true for your function). Moreover, the lower bound you find might not take on its minimum value at that same value of $x$, so even then the bound may not be sharp. In general AM-GM will not give you a sharp lower bound on the value of a function. – kccu Apr 04 '19 at 14:38
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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 $|\cos x|=|\sin x|$.

Oscar Lanzi
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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.

Michael Rozenberg
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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 $\sin2x=\pm1$.

farruhota
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