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Let $x,y,z\in \Bbb{R}^+$ such that $x^2+y^2+z^2+(x+y+z)^2\leq 4$. Prove that $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq 3.$$

As there are three fractions in the left side and a single term in the right side, I thought Cauchy-Schwarz might be of help. But from Cauchy-Schwarz, I got $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq \frac{\left(\sum_{cyc}\sqrt{xy+1}\right)^2}{x^2+y^2+z^2+(x+y+z)^2}.$$ Then we have to prove that $$\left(\sum_{cyc}\sqrt{xy+1}\right)^2\geq12.$$ But this doesn't seem to work. So, how to solve the problem?

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  • $\begingroup$ FYI In your application of CS, you forgot a square on the numerator, so your last conclusion doesn't follow. $\endgroup$
    – Calvin Lin
    Commented Aug 12, 2021 at 17:42
  • $\begingroup$ Please search before posting. This has been handled at AOPS. $\endgroup$ Commented Aug 13, 2021 at 4:55
  • $\begingroup$ Approach0 $\endgroup$ Commented Aug 13, 2021 at 4:56

2 Answers 2

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We have \begin{align*} \sum_{\mathrm{cyc}} \frac{xy + 1}{(x + y)^2} &\ge \sum_{\mathrm{cyc}} \frac{xy + \frac{x^2 + y^2 + z^2 + (x + y + z)^2}{4}}{(x + y)^2}\\ &= \sum_{\mathrm{cyc}} \frac{2xy + x^2 + y^2 + z^2 + xy + yz + zx}{2(x + y)^2}\\ &= \sum_{\mathrm{cyc}} \frac{(x + y)^2 + (y + z)(z + x)}{2(x + y)^2}\\ &= \frac{3}{2} + \sum_{\mathrm{cyc}} \frac{(y + z)(z + x)}{2(x + y)^2}\\ &\ge \frac32 + 3\sqrt[3]{\prod_{\mathrm{cyc}}\frac{(y + z)(z + x)}{2(x + y)^2}}\\ &= 3. \end{align*} We are done.

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$$\sum_{cyc}\frac{xy+1}{(x+y)^2}-3\geq\sum_{cyc}\frac{2xy+\sum\limits_{cyc}(x^2+xy)}{2(x+y)^2}-3=\frac{\sum\limits_{cyc}(x^2+3xy)\sum\limits_{cyc}(x^4-x^2y^2)}{2\prod\limits_{cyc}(x+y)^2}\geq0.$$

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