# show this inequality with $xy+yz+zx=3$

let $$x,y,z>0$$ and such $$xy+yz+zx=3$$,show that $$\dfrac{x}{x^3+y^2+1}+\dfrac{y}{y^3+z^2+1}+\dfrac{z}{z^3+x^2+1}\le 1$$

To prove this inequality,I want use following Cauchy-Schwarz inequality $$(x^3+y^2+1)(\frac{1}{x}+1+z^2)\ge (x+y+z)^2$$ $$\dfrac{x}{x^3+y^2+1}\le \dfrac{1+x+xz^2}{(x+y+z)^2}$$ we have $$\sum\dfrac{x}{x^3+y^2+1}\le\dfrac{3+x+y+z+xz^2+yx^2+zy^2}{(x+y+z)^2}$$ it suffices to prove that $$x+y+z+xz^2+yx^2+zy^2\le x^2+y^2+z^2+3$$

• Should the last inequality to prove be $x+y+z+xz^2+yx^2+zy^2\leq x^2+y^2+z^2+3$? – Yuta May 15 at 3:08
• @function sug: Your attempt only yields a sufficient condition, which can be shown to not always hold. For example, using $(x,y,z)=(6/5,1/5,69/35)$, you get $xy+yz+zx=3$, but $x+y+z+xz^2+yx^2+zy^2 > x^2+y^2+z^2+3$. – quasi May 15 at 3:49

By Muirhead $$\sum_{cyc}\frac{x}{x^3+y^2+1}\leq\sum_{cyc}\frac{x}{x^2+y^2+x}.$$ Thus, it's enough to prove that $$\sum_{cyc}\frac{x}{x^2+y^2+x}\leq1$$ or $$\sum_{cyc}\left(x^4y^2+x^4z^2+\frac{2}{3}x^2y^2z^2-x^3y-x^2yz-\frac{2}{3}xyz\right)\geq0$$ or $$\sum_{cyc}(3x^4y^2+3x^4z^2+2x^2y^2z^2-x^4y^2-x^4yz-x^3y^2z-x^3y^2z-x^3z^2y-x^2y^2z^2-2xyz)\geq0$$ or $$\sum_{cyc}(2x^4y^2+3x^4z^2-x^4yz-2x^3y^2z-x^3z^2y+x^2y^2z^2-2xyz)\geq0$$ or $$\sum_{cyc}(x^4z^2-x^3y^2z)+2\sum_{cyc}(x^4y^2+x^4z^2-2x^4yz)+$$ $$+xyz\sum_{cyc}(x^3-x^2y-x^2z+xyz)+2xyz\sum_{cyc}(x^3-1)\geq0,$$ which is true by AM-GM and Schur.