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Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Prove that $$\frac{2-2a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2} \leq \frac{9}{4}$$ I find when the equality occurs at $a=\frac{1}{\sqrt{15}},b=c=\sqrt{\frac{3}{5}}$ but I don't know how to prove it. I tried to use the equality $1+a^2=\left(a+b\right)\left(a+c\right)$ but it seems to be more difficult than the original one.

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There is also the following reasoning.

We need to prove that $$\frac{2(1-a^2)}{1+a^2}-2+\frac{1-b^2}{1+b^2}-1+\frac{1-c^2}{1+c^2}-1\leq\frac{9}{4}-4$$ or $$\frac{2a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2}\geq\frac{7}{8}.$$ Now, by C-S $$\frac{2a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2}=\frac{a^2}{\frac{1+a^2}{2}}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2}\geq$$ $$\geq\frac{(a+b+c)^2}{\frac{1+a^2}{2}+1+b^2+1+c^2}=\frac{2(a+b+c)^2}{a^2+2b^2+2c^2+5(ab+ac+bc)}.$$ Id est, it's enough to prove that $$16(a+b+c)^2\geq7(a^2+2b^2+2c^2+5(ab+ac+bc))$$ or $$9a^2-3(b+c)a+2b^2+2c^2-3bc\geq0,$$ for which it's enough to prove that $$(b+c)^2-4(2b^2+2c^2-3bc)\leq0$$ or $$(b-c)^2\geq0.$$ Done!

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  • $\begingroup$ Thank you very much. I got it. $\endgroup$ – RuaSun Mar 29 at 8:41
  • $\begingroup$ @RuaSun You are welcome! $\endgroup$ – Michael Rozenberg Mar 29 at 11:36
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Let $a=x$, $b=3y$ and $c=3z$.

Thus, after homogenization and full expanding we need to prove that: $$3x^2y+3x^2z+y^2x+3y^2z+z^2x+3z^2y\geq14xyz,$$ which is true by AM-GM: $$3x^2y+3x^2z+y^2x+3y^2z+z^2x+3z^2y\geq14\sqrt[14]{x^{6+6+1+1}y^{3+2+6+3}z^{3+3+2+6}}=14xyz.$$ Done!

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