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Prove that for positive numbers x and y the inequality $$ \sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x}{1+y}}+\sqrt{\dfrac{y}{1+x}}>2$$ is true.

I tried to use the inequality $$\dfrac{1}{\sqrt{ab}} \geq \dfrac{2}{a+b}.$$

For example, the first fraction turns to $$\sqrt{\dfrac{1}{(x+y)1}} \geq \dfrac{2}{x+y+1}.$$

The second fraction turns to $$\sqrt{\dfrac{x}{(1+y)1}} \geq \dfrac{2x}{y+2}.$$

The third fraction turns to $$\sqrt{\dfrac{y}{(1+x)1}} \geq \dfrac{2y}{x+2}.$$

So we have $$ \sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x}{1+y}}+\sqrt{\dfrac{y}{1+x}} \geq \dfrac{2}{x+y+1}+\dfrac{2x}{2+y}+\dfrac{2y}{2+x} = 2 \left( \dfrac{1}{x+y+1}+\dfrac{x}{2+y}+\dfrac{y}{2+x}\right)>2.$$ Is it enough to prove that multiplying by 2, we get an expression greater than two?

Any hint would help a lot, Thanks!

Alice Malinova
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  • Can you show how you applied the inequality for each of the 3 terms? – Calvin Lin May 27 '23 at 14:37
  • $\frac{2}{\frac{1}{x} + y + 1} ≠ \frac{2x}{2+y}$ – An_Elephant May 27 '23 at 14:45
  • @CalvinLin yes, I added this part, you can see it now –  Alice Malinova May 27 '23 at 15:08
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    @MichaelRozenberg: Why did you reopen the question? It looks like an exact duplicate of https://math.stackexchange.com/q/1003117/42969 to me (and at least one of your solutions posted here is quite similar to what you answered to the older question). – Martin R May 27 '23 at 16:19
  • @Martin R Your inequality has three variables, but the starting problem has two variables. They are really different problems. Maybe you found out that they are both inequalities? It's not enough for me. – Michael Rozenberg May 27 '23 at 16:25
  • So this is a special case of the other problem, and even you treat it as $\sum_{cyc}\sqrt{\frac{x}{y+z}}$ in your answer below, and the solutions are identical. – Martin R May 27 '23 at 16:27
  • @Martin R Yes, of course, I used math. It does not say that we need to delete any problems, where we use math. Don't agree. – Michael Rozenberg May 27 '23 at 16:31
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    @MichaelRozenberg Who said anything about deleting anything? Closing a question as a duplicate protects it from the kind of automatic deletion that occurs with other close reasons, and helps future users to see related content. I agree with Martin R that this is a duplicate question, and that it should be closed with a link to the duplicate. – Xander Henderson May 27 '23 at 16:34
  • @Xander Henderson♦ I don't agree with you. It's not duplicate. Also, you deleted very many "duplicates" like this. In my opinion it's very bad. – Michael Rozenberg May 27 '23 at 16:41
  • @AliceMalinova To clarify, can you show the second term too? I only got $2\sqrt{x} / (2+y)$, or $ 2x/(xy+x+1)$ like An_Elephant. – Calvin Lin May 27 '23 at 18:13
  • @CalvinLin I added, see it now. –  Alice Malinova May 27 '23 at 18:18
  • Can you write it out step by step? Please humor me and show all of the working. $\quad$ FWIW I still only get $ 2 \sqrt{x} / (2+y)$ using what I think is your approach – Calvin Lin May 27 '23 at 18:21

2 Answers2

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Let $1=z$.

Thus, by Holder and Schur we obtain: $$LHS=\sum_{cyc}\sqrt{\frac{x}{y+z}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{x}{y+z}}\right)^2\sum\limits_{cyc}x^2(y+z)}{\sum\limits_{cyc}x^2(y+z)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}x^2(y+z)}}>2.$$

Michael Rozenberg
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Another way. By AM-GM $$\sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x}{1+y}}+\sqrt{\dfrac{y}{1+x}}=\dfrac{1}{\sqrt{1(x+y)}}+\dfrac{x}{\sqrt{x(1+y)}}+\dfrac{y}{\sqrt{y(1+x)}}\geq$$ $$\geq\frac{2}{1+x+y}+\frac{2x}{x+1+y}+\frac{2y}{y+1+x}=2.$$ The equality occurs for $1=x+y$, $x=1+y$ and $y=1+x,$ which says that the equality does not occur.

Michael Rozenberg
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