prove $\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq 1$

Question

If $x$,$y$,$z$ are positive real numbers,Prove:$$\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq 1$$ Using this two inequality:

$\sum ^n_{i=1} \sqrt{a_ib_i}\leq\sqrt {ab} $ (we call it $A$ inequality)

$\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$ (we call it $B$ inequality)

which $a_i$ and $b_i$ are positive and and $b= \sum ^n_{i=1} b_i$,$a= \sum ^n_{i=1} a_i$.

Additional info: The question emphasizes in using inequalities $A$ and $B$.Beside them we can use AM-GM and Cauchy inequalities only.We are not allowed to use induction.And if you like,here you can see Prove of inequality B.

Things I have tried so far:

Using inequality $A$ i can re write question inequality as$$\sum \limits_{cyc} \frac{x}{2x+\sqrt{xy}}\leq 1$$

And I can't go further.I can't observer something that could lead me to using inequality $B$.

score 9 · Accepted Answer · edited Jan 28 '18 at 14:07

9

Use the fact $\sqrt{(x+y)(x+z)}\ge \sqrt{xy}+\sqrt{xz}$. Which is obvious after squaring and cancelling terms. Now our inequality becomes : $$\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\le \sum \limits_{cyc} \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=\sum \limits_{cyc} \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1$$

edited Jan 28 '18 at 14:07

Vee Hua Zhi

881

answered Jul 30 '14 at 14:15

shadow10

5,616

thanks for hint,i'm trying figuring out the part you write $\sum \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=1$.can you give me a hint on how you concluded that? – user2838619 Jul 30 '14 at 14:40
That was not a hint actually, its the full solution lol. Anyhoo I am editing it so you get a clearer idea. – shadow10 Jul 30 '14 at 15:10
How i missed a simple factoring like that.thanks. – user2838619 Jul 30 '14 at 15:19
So only inequality A is needed. – Ewan Delanoy Jul 30 '14 at 15:19
2

funny thing is Mathematica has serious problem with it – user2838619 Jul 30 '14 at 16:19

score 2 · Answer 2 · answered Nov 21 '18 at 03:07

By your work and by C-S we obtain: $$\sum_{cyc}\frac{x}{x+\sqrt{(x+y)(x+z)}}\leq\sum_{cyc}\frac{x}{2x+\sqrt{yz}}=\frac{3}{2}-\sum_{cyc}\left(\frac{1}{2}-\frac{x}{2x+\sqrt{yz}}\right)=$$ $$=\frac{3}{2}-\frac{1}{2}\sum_{cyc}\frac{\sqrt{yz}}{2x+\sqrt{yz}}=\frac{3}{2}-\frac{1}{2}\sum_{cyc}\frac{yz}{2x\sqrt{yz}+yz}\leq$$ $$\leq\frac{3}{2}-\frac{1}{2}\cdot\frac{\left(\sum\limits_{cyc}\sqrt{yz}\right)^2}{\sum\limits_{cyc}(2x\sqrt{yz}+yz)}=\frac{3}{2}-\frac{1}{2}=1.$$

score 1 · Answer 3 · answered Jun 30 '23 at 18:26

1

Another way.

By AM-GM we obtain: $$\sum_{cyc}\tfrac{x}{x+\sqrt{(x+y)(x+z)}}=\sum_{cyc}\tfrac{x\left(\sqrt{(x+y)(x+z)}-x\right)}{xy+xz+yz}\leq\sum_{cyc}\tfrac{x\left(\frac{1}{2}(x+y+x+z)-x\right)}{xy+xz+yz}=1.$$

answered Jun 30 '23 at 18:26

Michael Rozenberg

194,933

prove $\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq 1$

3 Answers3