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let $x,y,z>0$ and such $x+y+z=1$, show that $$(xy+yz+xz)\left(\dfrac{xy}{z^2+1}+\dfrac{yz}{x^2+1}+\dfrac{zx}{y^2+1}\right)\le\dfrac{1}{10}$$

my idea: $$\dfrac{xy}{z^2+1}=\dfrac{xy}{z^2+(x+y+z)^2}=\dfrac{xy}{2z^2+2xy+2yz+2xz+x^2+y}$$ Maybe this is old inequality,and It is said can use Cauchy-Schwarz inequality to solve it

Thank you

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we can suppose that $ x=\dfrac{a}{a+b+c} , y=\dfrac{b}{a+b+c} , z=\dfrac{c}{a+b+c} $ where $ a,b,c>0 $ are three real numbers.

suppose that $ s=a+b+c $

now $ inequality \Leftrightarrow (\sum{\dfrac{ab}{s^2}})(\sum{\dfrac{(\dfrac{ab}{s^2})}{(\dfrac{c^2+s^2}{s^2})}}) \le \dfrac{1}{10} $

so we must prove that $ \dfrac{(\sum ab )}{s^2}(\sum {\dfrac{ab}{c^2+s^2}}) \le \dfrac{1}{10} $

for $ a,b,c>0 $

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By C-S and AM-GM $$\sum_{cyc}\frac{xy}{z^2+1}=\sum_{cyc}\frac{xy}{2(x+z)(y+z)+x^2+y^2}\leq\sum_{cyc}\frac{xy}{(8+2)^2}\left(\frac{8^2}{2(x+z)(y+z)}+\frac{2^2}{x^2+y^2}\right)=$$ $$=\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}+\frac{1}{50}\sum_{cyc}\frac{2xy}{x^2+y^2}\leq\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}+\frac{3}{50}.$$ Thus, it remains to prove that $$\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}\leq\frac{(x+y+z)^2}{10(xy+xz+yz)}-\frac{3}{50}$$ or $$\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}\leq\frac{\sum\limits_{cyc}(5x^2+7xy)}{50(xy+xz+yz)}$$ or $$\sum\limits_{cyc}(5x^2+7xy)\prod_{cyc}(x+y)\geq16(xy+xz+yz)\sum_{cyc}(x^2y+x^2z)$$ or $$\sum_{cyc}(5x^4y+5x^4z-4x^3y^2-4x^3z^2-8x^3yz+6x^2y^2z)\geq0$$ or $$5\sum_{cyc}(x^4y+x^4z-x^3y^2-x^3z^2)+\sum_{cyc}(x^3y^2+x^3z^2-2x^3yz)-6xyz\sum_{cyc}(x^2-xy)\geq0$$ or $$\sum_{cyc}(x-y)^2(5xy(x+y)+z^3-3xyz)\geq0,$$ which is true by AM-GM: $$5xy(x+y)+z^3-3xyz\geq xy(x+y)+z^3-3xyz\geq$$ $$\geq2\sqrt{x^3y^3}+z^3\geq3\sqrt[3]{x^3y^3z^3}-3xyz=0.$$ Done!

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