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$x,y,z$ are positive real numbers, such that $x+y+z=3$ . prove that :

$$\sum_{\mathrm{cyc}}\frac{x}{x^3+y^2+z} \leq 1 $$

I tried many things , but I don't think any of those are worth of mentioning. However, I know problem can be solves using Cauchy–Schwarz inequality.

Please, share your ideas. Thanks.

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    $\begingroup$ Since this is contest-math, please indicate a source for the problem. $\endgroup$
    – Aphelli
    Sep 21, 2021 at 17:13
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    $\begingroup$ There is an answer here but it uses Holder's inequality. Read this as well. I think this is safely not a current contest question because of the similarities to the questions I've just presented. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Sep 21, 2021 at 17:16
  • $\begingroup$ Unfortunately , I don't know where the problem comes from. It was on IZHO TST in my city few years ago, but I don't think they came up with it. $\endgroup$
    – mate zhorzholiani
    Sep 21, 2021 at 17:20
  • $\begingroup$ @TeresaLisbon Thank you very much! $\endgroup$
    – mate zhorzholiani
    Sep 21, 2021 at 17:21
  • $\begingroup$ You are welcome! If you are able to write an answer using only AM-GM and using the hints given, it will be useful. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Sep 21, 2021 at 17:22

1 Answer 1

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By C-S $$\sum_{cyc}\frac{x}{x^3+y^2+z}=\sum_{cyc}\frac{x\left(\frac{1}{x}+1+z\right)}{(x^3+y^2+z)\left(\frac{1}{x}+1+z\right)}\leq\sum_{cyc}\frac{(1+x+xz)}{(x+y+z)^2}\leq1.$$ Can you end it now?

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