## Inequality with a,b,c sides of a triangle

For discussing Olympiad Level Algebra (and Inequality) problems
Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Inequality with a,b,c sides of a triangle

Let $a,b$ and $c$ be sides of a triangle. Prove that:
$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3$

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: Inequality with a,b,c sides of a triangle

$\sqrt{b}+\sqrt{c}-\sqrt{a}=x,\sqrt{c}+\sqrt{a}-\sqrt{b}=y,\sqrt{a}+\sqrt{b}-\sqrt{c}=z$

$$\sum_{cyc} \frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}} = \sum_{cyc} \sqrt{1- \frac{(x-y)(x-z)}{2x^2}} \le \sum_{cyc} 1-\frac{(x-y)(x-z)}{4x^2} = 3 - \frac{1}{4} \sum_{cyc} x^{-2}(x-y)(x-z)$$

So, we need to prove $$\sum_{cyc} x^{-2}(x-y)(x-z) \ge 0$$ which is obviously true by Schur's ineq.
This was freedom. Losing all hope was freedom.

Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Re: Inequality with a,b,c sides of a triangle

Sorry Atonu, but I don't understand some passages...
Why exactly

$$\sum_{cyc} \frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}= \sum_{cyc} \sqrt{1- \frac{(x-y)(x-z)}{2x^2}} \le \sum_{cyc} 1-\frac{(x-y)(x-z)}{4x^2} = 3 - \frac{1}{4} \sum_{cyc} x^{-2}(x-y)(x-z)$$ ?

And how do you use exactly Shur's inequality to show that $$\sum_{cyc} x^{-2}(x-y)(x-z) \ge 0$$ ?

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: Inequality with a,b,c sides of a triangle

Katy729 wrote:
Fri Apr 20, 2018 10:24 pm
Sorry Atonu, but I don't understand some passages...
Why exactly

$\sum_{cyc} \frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}= \sum_{cyc} \sqrt{1- \frac{(x-y)(x-z)}{2x^2}} \le \sum_{cyc} (1-\frac{(x-y)(x-z)}{4x^2}) = 3 - \frac{1}{4} \sum_{cyc} x^{-2}(x-y)(x-z)$ ?

And how do you use exactly Shur's inequality to show that $$\sum_{cyc} x^{-2}(x-y)(x-z) \ge 0$$ ?
$\sqrt{b}+\sqrt{c}-\sqrt{a}=x \Rightarrow b+c-a=x^2- 2a-2\sqrt{bc}+2\sqrt{ca}+2\sqrt{ab}= x^2 - \frac{1}{2}(x-y)(x-z)$
$\therefore \frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}} = \sqrt{\frac{x^2 - \frac{1}{2}(x-y)(x-z)}{x^2}}=\sqrt{1 - \frac{(x-y)(x-z)}{2x^2}}$
Let $\frac{(x-y)(x-z)}{x^2}=n$. Now, all I wanted to do is flip the root. That is, $\sqrt{1-\frac{n}{2}}=\sqrt{1-2\frac{n}{4}} \le \sqrt{1-2\frac{n}{4}+\frac{n^2}{16}} = 1- \frac{n}{4}$.

And, Schur's ineq states that $\sum_{cyc}x^n(x-y)(x-z) \ge 0$ for positive real $x,y,z$ and integer $n$.
This was freedom. Losing all hope was freedom.

Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Re: Inequality with a,b,c sides of a triangle

Thanks Atonu, now is clear!
Only a doubt, can you put a link where is wrote the property that you say?
Because I found that the exponent ($n$) must be positive, yes in your cases the numbers are positive but where is wrote? I saw here but I didn't find the property that you say...

https://en.m.wikipedia.org/wiki/Schur%27s_inequality

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: Inequality with a,b,c sides of a triangle

Katy729 wrote:
Sat Apr 21, 2018 10:26 pm
Thanks Atonu, now is clear!
Only a doubt, can you put a link where is wrote the property that you say?
Because I found that the exponent ($n$) must be positive, yes in your cases the numbers are positive but where is wrote? I saw here but I didn't find the property that you say...

https://en.m.wikipedia.org/wiki/Schur%27s_inequality
Okay, the exponent must not be positive. One can easily prove that if the exponent is negative, the inequality still holds.
And, you can see $x,y,z$ are obv positive. Because, $b+c>a \Rightarrow b+c+2\sqrt{bc}>a \Rightarrow (\sqrt{b}+\sqrt{c})^2>(\sqrt{a})^2 \Rightarrow \sqrt{b}+\sqrt{c}-\sqrt{a}>0$
I think now it is clear.
This was freedom. Losing all hope was freedom.

Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Re: Inequality with a,b,c sides of a triangle

Yes, now understand. Very gentle Atonu!

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm