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Help me to find the maximum value of $T$ with $x, y, z \in \Bbb{R_+}$ $$T=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$$ Thanks :D

sirfoga
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user146673
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  • It's symmetric in the three variables. If you know there's a unique maximum, I believe you can use that symmetry to your advantage. – jdc Apr 29 '14 at 17:12
  • I would see @jdc 's comment and raise him the use of monotonicity along those lines of symmetry – dcs24 Apr 29 '14 at 17:47
  • I call. [and some more characters] – jdc Apr 30 '14 at 14:41

1 Answers1

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Hint: first let $$\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}=a\\ \frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}=b\\ \frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}=c$$then using HM-GM-AM-QM Inequalities you get $$\frac{\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}}{3} = \frac{a+b+c}{3} \le \sqrt{\frac{a^2+b^2+c^2}{3}}$$ Moreover the maximum of $\frac{a+b+c}{3}$ is obtained if and only if $a=b=c$ and $\frac{a+b+c}{3} = \sqrt{\frac{a^2+b^2+c^2}{3}}$. Finally you need to solve \begin{cases} \frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}=\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}\\ \frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}=\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}\\ \frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3} \end{cases} But since it's a symmetrical system, the solution is $x=y=z$, thus $$\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3} = \frac{x^3x^4x^3}{(x^4+x^4)(x^2+x^2)^3}=\frac{1}{16}$$ Finally the maximum possible value of $T$ is $$T=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}=3\frac{x^3x^4x^3}{(x^4+x^4)(x^2+x^2)^3}=3\frac{1}{16}=\frac{3}{16}$$

sirfoga
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  • Well, I couldn't see a proof that disproved my solution...anyway if you substitute the solution $(x,x,x)$ in the symmetrical system, you can verify that it is a solution... – sirfoga Apr 29 '14 at 18:08
  • Well, I am not clear how your system is justified, for e.g. we can write an inequality e.g. $a+b+c \le 3 \max(a, b, c)$ with equality iff $a=b=c$, or a host of other such inequalities - why do you think all/any of these lead to the absolute maximum of $T$? – Macavity Apr 29 '14 at 18:20
  • Perhaps I should add that you can also show the inequality $a+b+c \ge 3\sqrt[3]{abc}$ with equality iff $a=b=c$, and by your argument conclude this must be a minimum also! and hence $T$ is a constant!! You see the contradiction? – Macavity Apr 29 '14 at 18:25