Weird Inequality, Weird Condition

I began exactly like Daniel did with the same substitution. Hence, substituting, $\displaystyle x = \csc \theta \text{ and } y = \sec \theta$ . Therefore the expression changes as,

$\displaystyle \frac{ x^3y^3 \text{ } + \text{ }x^2y \text{ }+\text{ } xy^2 \text{ } + \text{ } 1}{x^3y^3} = 1\text{ } + \text{ } \sin ^2 \theta \cos \theta \text{ } + \text{ } \cos ^2 \theta \sin \theta \text{ } + \text{ } \sin ^3 \theta \cos^ 3 \theta$

Let the expression whose range we need to find out be $\varepsilon (\theta )$ .

$\displaystyle \Rightarrow \varepsilon (\theta ) = (1\text{ } + \text{ } \sin ^2 \theta \cos \theta )( 1\text{ } + \text{ } \cos ^2 \theta \sin \theta )$

Basically we need to find the extremums of $\varepsilon (\theta )$ . Note that in expression $\varepsilon (\theta )$ the $\sin \theta$ and $\cos \theta$ terms are interchangeable. Hence, $\displaystyle \varepsilon (\theta ) = \varepsilon (\frac{(4n +1)\pi }{2} - \theta ) \text{ }\forall \text{ } n \text{ } \in \text{ } \mathbb{Z}$ .

Consider an atleast twice differentiable function $f(x)$ satisfying the property $\displaystyle f(x) = f(k-x)$ for some real constant $k$ . Differentaiting, we get, $\displaystyle f'(x) + f'(k-x) = 0 \Rightarrow f'(\frac{k}{2} ) = 0$ . Hence such a function will have an extremum at $\displaystyle x = \frac{k}{2}$ .

Similarly, for our expression $\varepsilon (\theta )$ , we will have extremum at $\displaystyle \frac{(4n + 1)\pi }{4} \forall \text{ } n \text{ } \in \text{ } \mathbb{Z} \text{ i.e. } \frac{\pi }{4} \text{, } \frac{5\pi }{4} \text{, } \dots$

Now it is easy to realize that the maximum and minimum at $\displaystyle \theta = \frac{\pi }{4} \text{ and } \theta = \frac{5\pi }{4}$ respectively. (These are the first values, there would be more values owing to the periodicity of the trigonometric functions. )

Hence, $\displaystyle M = \frac{9 + 4\sqrt{2} }{8} \text{ and } m = \frac{9 - 4\sqrt{2}}{8}$ .

Therefore, $\displaystyle \lfloor 1000 \{M + m\} \rfloor = \boxed{250}$

SOLUTION FOR MINIMUM IS NOT CORRECT. I FORGOT TO TAKE $\pm$ WHEN TAKING THE SQUARE ROOT.

We want to first manipulate the condition to something easier. So:

$\begin{aligned}x^2+y^2=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2&\implies x^2+y^2=\left(\dfrac{x^2+y^2}{xy}\right)^2\\ &\implies 1=\dfrac{x^2+y^2}{x^2y^2}\\ &\implies 1=\dfrac{1}{x^2}+\dfrac{1}{y^2}\end{aligned}$

This is similar to the trigonometric identity $\sin^2\theta+\cos^2\theta=1$ , except the fractions are in reverse. This makes us realize if we substitute $x=\csc\theta$ and $y=\sec \theta$ then it satisfies the original condition.

Now let's simplify the expression we want to find the maximum and minimum of. We see that $\dfrac{x^3y^3+x^2y+xy^2+1}{x^3y^3}=1+\dfrac{1}{x^2y}+\dfrac{1}{xy^2}+\dfrac{1}{x^3y^3}$

Substituting our trig formulas in, we want to find the max and min of $1+\dfrac{1}{\csc^2\theta\sec\theta}+\dfrac{1}{\csc\theta\sec^2\theta}+\dfrac{1}{\csc^3\theta\sec^3\theta}$

Simplifying: $\begin{aligned}1+\dfrac{1}{\csc^2\theta\sec\theta}+\dfrac{1}{\csc\theta\sec^2\theta}+\dfrac{1}{\csc^3\theta\sec^3\theta}&= 1+\sin^2\theta\cos\theta+\sin\theta\cos^2\theta+\sin^3\theta\cos^3\theta\\ &= 1+\sin\theta\cos\theta(\sin\theta+\cos\theta)+(\sin\theta\cos\theta)^3\\ &= 1+\dfrac{1}{2}\sin2\theta\sqrt{1+\sin 2\theta}+\dfrac{1}{8}\sin^32\theta\\ &= 1+\dfrac{1}{2}n\sqrt{n+1}+\dfrac{1}{8}n^3\end{aligned}$

where the last step was substituting $n=\sin 2\theta$

We then must find the maximum and minimum of $f(n)=1+\dfrac{1}{2}n\sqrt{n+1}+\dfrac{1}{8}n^3$ where $n\in [-1,1]$ .

First, off the maximum. We find that, after messy second derivatives, that the function $f(n)$ is convex; thus, the maximum is at one of its endpoints: $n=-1,1$ . Checking both, we see that the corresponding maximums are $f(-1)=\dfrac{7}{8}$ and $f(1)=\dfrac{9}{8}+\dfrac{\sqrt{2}}{2}$ .

Clearly, $f(-1) < f(1)$ so the maximum is $\boxed{M=\dfrac{9}{8}+\dfrac{\sqrt{2}}{2}}$

Now for the harder part: the minimum. Unfortunately, there doesn't seem to be an easy way. Since $f(n)$ is convex, if we set $f'(n)=0$ , then we are guaranteed to find the minimum. After some messy deriving, we see that $f'(n)=\dfrac{3\sqrt{n+1}n^2+6n+4}{8\sqrt{n+1}}=0$

We can get rid of the denominator: $3\sqrt{n+1}n^2+6n+4=0$

The root of this is approximated to be (sorry!) $n\approx -0.809668$ and plugging this back in shows that the minimum is $m=f(-0.8097)\approx \boxed{0.757}$

Thus, our answer is $0.757+\dfrac{9}{8}+\dfrac{\sqrt{2}}{2}=2.\boxed{589}$

---

Equality case for maximum is $(x,y)=(\sqrt{2},\sqrt{2})$

Equality case for minimum is $(x,y)\approx (1.123,-2.200)$

x²+y²=(x²+y²)²/(xy)² so 1=(x²+y²)/(x²y²)=1/y²+1/x² Now replace 1/y=sin(t),1/x=cos(t) So y=1/cos(t),x=1/sin(t) f(x,y)=f(t)=(1+1/(sin²(t)cos(t))+1/(cos²(t)sin(t))+1/(sin³(t)cos³(t)))/(1/(sin³(t)cos³(t))=1+sin(t)cos(t) (sin(t)+cos(t))+(sin(t)cos(t))³ Notist that sin(t)cos(t)=0.5sin(2t) And max is when sin(2t)=1,i.e. 2t=π/2+2kπ,so t=π/4+kπ So sin(t)=cos(t)=±1/√2 Similiar sin(t)+cos(t)=√2 (1/√2sin(t)+1/√2cos(t))=√2sin(t+π/4),so max is when t+π/4=π/2+2kπ i.e. t=π/4+2kπ

So cos(t)=sin(t)=1/√2 So in cae t=π/4+2kπ max is achieved I.e.M=1+√2/2+(1/8)=(9+4√2)/8 Similiar notist that min of is when sin(x)=cos(x)=-1/√2 So m=1-√2/2+1/8=(9-4√2)/8 So m+M=18/8 So [1000{m+M}]=[1000*2/8]=250 Notist that f(t)=(1+sin(t)cos²(t))(1+cos(t)sin²(t))

I also arrived at 250, but with almost zero effort. I'll admit my solution is not elegant or anything close, but here's what I did. Simplified the first relation to arrive at exactly what Daniel did, $\frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ { y }^{ 2 } } =1$ From there, x and y could be sqrt(2) or -sqrt(2). [This was just lucky. ] So applying both the values in the expression, the values were (9-4 sqrt2)/8 and (9+4 sqrt2)/8

Adding them I got 9/8 or 2.25, and taking the fractional part and times 1000, finally 250. This is obviously not the ideal way to solve it, but I think I got lucky here.