Find the maximum possible value

Question

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

Answer 1

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 $$T=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}=3\frac{x^3x^4x^3}{(x^4+x^4)(x^2+x^2)^3}=3\frac{1}{16}=\frac{3}{16}$$