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For $x, y, z > 0$, prove that $$\sqrt{\dfrac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\dfrac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}>1.$$

I tried to use the AM-GM inequality as in the previous question $$\dfrac{x}{\sqrt{x\left(x+y\right)\left(1+y\right)}}+\dfrac{y}{\sqrt{y\left(x+y\right)\left(1+x\right)}}+\dfrac{xy}{\sqrt{xy\left(1+x\right)\left(1+y\right)}} \geq \dfrac{2x}{2x+xy+y+y^2}+\dfrac{2y}{2y+xy+x+x^2}+\dfrac{2xy}{1+x+y+2xy}=2 \cdot \left( \dfrac{x}{2x+xy+y+y^2}+\dfrac{y}{2y+xy+x+x^2}+\dfrac{xy}{1+x+y+2xy}\right).$$

The question again is it enough to prove the inequality this way?

River Li
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Alice Malinova
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    Please add a link if you refer to a previous question. – Martin R May 27 '23 at 16:01
  • @MartinR please https://math.stackexchange.com/questions/4707567/prove-that-for-positive-numbers-x-and-y-the-inequality-is-true –  Alice Malinova May 27 '23 at 16:16
  • Again, do you mind showing your AM-GM steps? $\quad$ FWIW You don't seem to be trying to "maintain equality" throughout, so that suggests that you're unlikely to prove the inequality. – Calvin Lin May 27 '23 at 18:19

3 Answers3

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By AM-GM, we have \begin{align*} &\sqrt{\frac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\frac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\\[6pt] ={}& \frac{2x}{2\sqrt{(x + y) \cdot x(1 + y)}} + \frac{2y}{2\sqrt{(x + y)\cdot y(1 + x)}} + \frac{2xy}{2\sqrt{x(1 + y) \cdot y(1+x)}}\\[6pt] \ge{}& \frac{2x}{x+ y + x(1 + y)} + \frac{2y}{x + y + y(1 + x)} + \frac{2xy}{x(1+y) + y(1+x)}\\[6pt] ={}& \frac{2x}{2x + y + xy} + \frac{2y}{x + 2y + xy} + \frac{2xy}{x + y + 2xy}\\[6pt] >{}& \frac{2x}{2x + 2y + 2xy} + \frac{2y}{2x + 2y + 2xy} + \frac{2xy}{2x + 2y + 2xy} \\[6pt] ={}& 1. \end{align*}

We are done.

River Li
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  • great solution! I wonder how did you come up with this :D ps in the second fraction it should be 2y in the numerator –  Alice Malinova May 28 '23 at 11:33
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    @AliceMalinova The idea comes from your previous question. In detail, in your previous question, we use AM-GM to get $$\sqrt{\frac{1}{x+y}}+\sqrt{\frac{x}{1+y}}+\sqrt{\frac{y}{1+x}} \ge \frac{2}{x + y + 1} + \frac{2x}{x + 1 + y} + \frac{2y}{y + 1 + x} = 2$$. In the spirit of this idea, we hope to construct the appropriate AM-GM. Edited the typo. Thanks. – River Li May 28 '23 at 12:13
  • yeah, I tried to do it by analogy, but something went wrong. You still need to correctly notice what to multiply in the denominators, so that later everything turns out to be beautifully the same and reduced. Great job! –  Alice Malinova May 28 '23 at 12:20
  • @AliceMalinova Are there more problems similar (to your previous and current)? – River Li May 28 '23 at 12:28
  • a lot! now I'm struggling with this https://math.stackexchange.com/questions/4708104/prove-2a2b2c2-frac43-sum-limits-mathrmcyc-frac1a21-geq –  Alice Malinova May 28 '23 at 12:46
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For positive variables let $1=z$.

Thus, by Holder $$LHS^2=\frac{\left(\sum\limits_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\right)^2\sum\limits_{cyc}x^2y^2(x+z)(y+z)}{\sum\limits_{cyc}x^2y^2(x+z)(y+z)}\geq\frac{(xy+xz+yz)^3}{\sum\limits_{cyc}x^2y^2(x+z)(y+z)}>1.$$

Michael Rozenberg
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  • nice solution, thanks! but what can you say about applying am-gm inequality like I tried to do this, is it possible at all? –  Alice Malinova May 27 '23 at 16:52
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    @Alice Malinova We can say that in my second solution I used AM-GM: $2\sum\limits_{cyc}\sqrt{x^2yx(x+y)(x+z)}\geq4\sum\limits_{cyc}xyz>\frac{2}{3}\sum\limits_{cyc}xyz.$ – Michael Rozenberg May 27 '23 at 16:58
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Another way.

For positive variables let $1=z$.

Thus, $$LHS^2=\left(\sum\limits_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\right)^2=\frac{\left(\sum\limits_{cyc}\sqrt{xy(x+y)}\right)^2}{\prod\limits_{cyc}(x+y)}=$$ $$=\frac{\sum\limits_{cyc}\left(x^2y+x^2z+2\sqrt{x^2yz(x+y)(x+z)}\right)}{\prod\limits_{cyc}(x+y)}>\frac{\sum\limits_{cyc}\left(x^2y+x^2z+\frac{2}{3}xyz\right)}{\prod\limits_{cyc}(x+y)}=1.$$

Michael Rozenberg
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