Let $x,y,z\ge 0: xy+yz+zx=1.$ Find the maximum value $$T={\frac {1-2yz}{3\,{y}^{2}+3\,{z}^{2}+{x}^{2}}}+{\frac {1-2zx }{3\,{z}^{2}+3\,{x}^{2}+{y}^{2}}}+{\frac {1-2xy}{3\,{x}^{2}+3\,{y}^ {2}+{z}^{2}}}.$$ I thought the max is $\dfrac{3}{7}$ when $x=y=z=\dfrac{\sqrt{3}}{3}$
I tried to use Cauchy-Schwarz $$\frac{6-12yz}{3.2(x^2+y^2+z^2)+2.3(y^2+z^2)}\le \frac{6-12yz}{25}\left(\frac{3}{2(x^2+y^2+z^2)}+\frac{2}{3(y^2+z^2)}\right)$$ The remain is proving $$\frac{9}{25(x^2+y^2+z^2)}+\frac{4}{25}\sum_{cyc}\frac{1-2yz}{y^2+z^2}\le \dfrac{3}{7}$$ But I don't know how to prove it. Could you please give me a hint to kill the last inequality ? Thank you very much.
You inequality is true for any reals $x$, $y$ and $z$ such that $x^2+y^2+z^2\neq0$.
Indeed, after homogenization we need to prove that: $$\sum_{sym}(13.5x^6-63x^5y+117x^4y^2+10.5x^4yz-49x^3y^3-70x^3y^2z+41x^2y^2z^2)\geq0.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Thus, a coefficient before $w^6$ in the left side it's: $$27\cdot3+63\cdot3-117\cdot3+21\cdot3-98\cdot3+70\cdot3-41\cdot6=144$$ Thus, the inequality $$\sum_{sym}(13.5x^6-63x^5y+117x^4y^2+10.5x^4yz-49x^3y^3-70x^3y^2z+41x^2y^2z^2)\geq$$ $$\geq\frac{16}{225}\left(\sum_{cyc}(3x^3-5x^2y-5x^2z+7xyz)\right)^3$$ is a linear inequality of $w^3$.
Indeed, a coefficient before $w^6$ in the right side it's: $$\frac{16}{225}(3\cdot3+5\cdot3+7\cdot3)^2=144.$$ Id est, by $uvw$ it's enough to prove the last inequality for $y=z=1$(for $y=z=0$ it's obvious), which gives $$(x-1)^2(3x-4)^2(659x^2+32x+884)\geq0.$$ About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791