Given $x,y,z>0$ such that $xyz=1$ Prove that: $\frac{x^7}{x^8+1}+\frac{y^7}{y^8+1}+\frac{z^7}{z^8+1}\leq \frac{3}{2}$ P/s: I tried to solve it by AM-HM but I failed
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Substitute $$x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$$ – Dr. Sonnhard Graubner Jan 08 '18 at 15:02
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Yes, and BW, but it's very ugly. I checked it. It's true. – Michael Rozenberg Jan 08 '18 at 15:03
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@MichaelRozenberg What is BW? – José Carlos Santos Jan 08 '18 at 15:10
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@MichaelRozenberg Never mind. I found it. – José Carlos Santos Jan 08 '18 at 15:16
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since \begin{align*} &\frac34-\frac34\cdot\frac1{x^{12}+x^6+1}-\frac{x^7}{x^8+1}\\ ={}&\frac{x^6(x-1)^2(3x^{12}+2x^{11}+x^{10}-x^8-2x^7-2x^5-x^4+x^2+2x+3)}{4(x^{12}+x^6+1)(x^8+1)} \end{align*} and \begin{align*} 2x^{11}+2x&\geqslant2x^7+2x^5,\\ x^{10}+x^2&\geqslant x^8+x^4, \end{align*} we get $$\frac{x^7}{x^8+1}\leqslant\frac34-\frac34\cdot\frac1{x^{12}+x^6+1},$$ so, just need to prove $$\frac94-\frac34\sum\frac1{x^{12}+x^6+1}\leqslant\frac32,$$ or $$\sum\frac1{x^{12}+x^6+1}\geqslant1,$$ or equivalent to: $a$, $b$, $c>0$, $abc=1$, prove that $$\sum\frac1{a^2+a+1}\geqslant1,$$ which is well-known.
kuing
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