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Let $x, y, z > 0$. Prove that $$\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1.$$

I encountered this problem today and thought that it was straightforward, and to be honest it still might be I just can't figure it out for some reason. I approached this problem using the famous Cauchy-Schwarz inequality. I applied the inequality to every denominator. For example for the first fraction: $$(1+xy+xz)(1 + \frac{1}{x}+ \frac{1}{x}) \ge (1+y+z)^2$$ After doing this for every fraction I switched the denominator for every one of them with the left side of the Cauchy-Schwarz inequality. Reducing each fraction we are left with $$\frac{1}{1+2/x} + \frac{1}{1+2/y} + \frac{1}{1+2/z} \ge 1 $$ I am not sure how to continue from here. I tried to use Titu's lemma but it didn't work. I will be very happy if someone can help me out here.

Edit: My Cauchy was wrong.Thanks to all who contributed :)

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    $\begingroup$ Solution on AoPS: artofproblemsolving.com/community/c6h331122p1771811 $\endgroup$
    – Martin R
    Commented Sep 24, 2022 at 15:53
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    $\begingroup$ Check your application of CS inequality on the first fraction - it isn't correct. Correct it and you'll find the rest a breeze. $\endgroup$
    – Macavity
    Commented Sep 25, 2022 at 4:37

2 Answers 2

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Hint :

Can you show ? For $a,b,c>0$

$$\frac{1+ba+bc}{(1+a+c)^{2}}-\frac{b^{2}+ba+bc}{(b+a+c)^{2}}=\frac{\left(b-1\right)^{2}\left(a+c\right)}{\left(a+c+1\right)^{2}\left(a+b+c\right)}\geq 0$$

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  • $\begingroup$ The rest is smooth. $\endgroup$ Commented Sep 25, 2022 at 10:16
  • $\begingroup$ @SuzuHirose The rest is even obvious ^^. $\endgroup$ Commented Sep 25, 2022 at 10:25
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By the Cauchy-Schwarz inequality, $$ (1+xy+xz)(1+y/x+z/x) \\ =\left(1,\sqrt{xy},\sqrt{xz}\right)^2 \left(1,\sqrt{ y/x},\sqrt{z/x}\right)^2\geq(1+y+z)^2 $$ which we can rearrange to $$ {1+xy+xz\over(1+y+z)^2}\geq{x\over x+y+z}. $$ The rest is smooth.

Credit

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