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Find all positive reals $x,y \in \mathbb{R}^+$ satisfying: $$\frac{x^9}{y} + \frac{y^9}{x} = 10-\frac{8}{xy}$$

Since this involves higher exponents I am unable to tackle this problem. Please help me.

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    I don't think this qualifies as a [Diophantine equation](https://en.m.wikipedia.org/wiki/Diophantine_equation). – 2017-01-16
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    Oh! I have now known that a Diophantine has only integer solutions but it involves reals. – 2017-01-16
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    Well, if you multiply it by $xy$ you will get a polynomial equation in $2$ variables of degree $10$. Generally there is no algorithm to solve such equations. But in that particular case? Who knows, doesn't look easy. – 2017-01-16
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    It looks cleaner as $x^{10}+y^{10}=10xy-8$ but I am not sure that is progress. – 2017-01-16
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    How about you tell us some story about the equation? Where did you get it from? Perhaps there's some hint in it. – 2017-01-16
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    $(x,y) = (1,1)$ is a solution. – 2017-01-16
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    A desmos graph https://www.desmos.com/calculator/timng19bz2 reveals that the only solutions are $(1,1)$ and $(-1,-1)$ but how to prove that? – 2017-01-16
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    @JohnWaylandBales I already tried the graph way – 2017-01-16
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    @freakish I came across this while solving a reference book – 2017-01-16

2 Answers 2

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Hint: Write it as:

$x^{10}+y^{10}=10xy-8$

Then use AM-GM:

$10xy-5=x^{10}+y^{10}+1+1+1\ge5\sqrt[5]{x^{10}y^{10}}=5x^2y^2$

Which gives $5(xy-1)^2\le0$, so $xy=1$

Then initial equation can be rewritten as $x^{10}+\frac{1}{x^{10}}=2$

Which by AM-GM again, (or by writing it as a $(x^{10}-1)^2=0$) has the solution, $x^{10}=1$, so $x=\pm1$

So $x=1,y=1$, or $x=-1,y=-1$

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    Wow! I never thought of using the AM-GM. What a brain! Thank for the answer. – 2017-01-16
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Just to give a non-AM-GM answer, let $u=xy$, which must be positive in order for $x^{10}+y^{10}=10xy-8$ to have a solution, and note that $x^{10}+y^{10}=(x^5-y^5)^2+2(xy)^5$. Thus

$$\begin{align}x^{10}+y^{10}-10xy+8 &=(x^5-y^5)^2+2(u^5-5u+4)\\ &=(x^5-y^5)^2+2(u-1)(u^4+u^3+u^2+u-4)\\ &=(x^5-y^5)^2+2(u-1)^2(u^3+2u^2+3u+4)\\ &\ge0 \end{align}$$

with equality if and only if $x^5=y^5$ and $xy=u=1$ (since $u\gt0$ implies $u^3+2u^2+3u+4\gt0$). From this we see that $x=y=\pm1$ are the only possibilities. In particular, $(x,y)=(1,1)$ is the only solution with $x,y\in\mathbb{R}^+$.

Remark: I showed the factorization of $u^5-5u+4$ in two steps for ease of checking.