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If $x,y,z$ are positive real numbers that $xyz=1$ , Prove

a) $\frac{xy}{x^8+xy+y^8}+\frac{yz}{y^8+yz+z^8}+\frac{zx}{z^8+zx+x^8}\leq1$

b)$\frac{(xy)^7}{x^8+(xy)^7+y^8}+\frac{(yz)^7}{y^8+(yz)^7+z^8}+\frac{(zx)^7}{z^8+(zx)^7+x^8}\leq1$

Additional info: We must use the following inequality for proving$$x^8+y^8\geq x^7y+y^7x\geq x^6y^2+y^6x^2\geq...\geq2x^4y^4$$

and more general this inequality $$a^{m+n}+b^{n+m}\geq a^nb^m+a^mb^n$$

Things I have tried so far: For Proving a) i can write $$\frac{xy}{x^8+xy+y^8}\leq\frac{xy}{x^7y+xy+y^7x}=\frac{1}{x^6+y^6+1}$$

And continuing this$$\frac{1}{x^6+y^6+1}\leq \frac{1}{x^4y^2+y^4x^2+x^2y^2z^2}=\frac{1}{x^2y^2(x^2+y^2+z^2)}=\frac{1}{\frac{1}{z^2}(x^2+y^2+z^2)}=\frac{z^2}{x^2+y^2+z^2}$$

By doing these for two other fractions we can prove inequality.

for part b, i tried to do same thing but I was unsuccessful. for example i end up with these inequality $\frac{(xy)^7}{x^8+(xy)^7+y^8}\leq \frac{x^3y^3}{2+x^2y^2}$ which is not useful for proving the problem.

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user2838619
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1 Answers1

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After the first step it will become $\frac{(xy)^6}{x^6+y^6+(xy)^6}=\frac{1}{\frac{1}{y^6}+\frac{1}{x^6}+1}$ After set as I said it will be $\frac{1}{a^6+b^6+1}$ Now you just need to continue what you do in a)

user169301
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  • thanks for your hint.could you please be more specific about the second part where you said "set $a=\frac{1}{x}$ ..." ?I did what you said and it just made it more complicated for me. – user2838619 Aug 13 '14 at 07:07