Inequality with a,b,c sides of a triangle
Inequality with a,b,c sides of a triangle
Let $a,b$ and $c$ be sides of a triangle. Prove that:
$\frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}}+\frac{\sqrt{c+ab}}{\sqrt{c}+\sqrt{a}\sqrt{b}}+\frac{\sqrt{a+bc}}{\sqrt{a}+\sqrt{b}\sqrt{c}}\leq 3$
$\frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}}+\frac{\sqrt{c+ab}}{\sqrt{c}+\sqrt{a}\sqrt{b}}+\frac{\sqrt{a+bc}}{\sqrt{a}+\sqrt{b}\sqrt{c}}\leq 3$
 Atonu Roy Chowdhury
 Posts: 64
 Joined: Fri Aug 05, 2016 7:57 pm
 Location: Chittagong, Bangladesh
Re: Inequality with a,b,c sides of a triangle
$\sqrt{b}+\sqrt{c}\sqrt{a}=x,\sqrt{c}+\sqrt{a}\sqrt{b}=y,\sqrt{a}+\sqrt{b}\sqrt{c}=z$
$$\sum_{cyc} \frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}} = \sum_{cyc} \sqrt{1 \frac{(xy)(xz)}{2x^2}} \le \sum_{cyc} 1\frac{(xy)(xz)}{4x^2} = 3  \frac{1}{4} \sum_{cyc} x^{2}(xy)(xz) $$
So, we need to prove $$\sum_{cyc} x^{2}(xy)(xz) \ge 0$$ which is obviously true by Schur's ineq.
$$\sum_{cyc} \frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}} = \sum_{cyc} \sqrt{1 \frac{(xy)(xz)}{2x^2}} \le \sum_{cyc} 1\frac{(xy)(xz)}{4x^2} = 3  \frac{1}{4} \sum_{cyc} x^{2}(xy)(xz) $$
So, we need to prove $$\sum_{cyc} x^{2}(xy)(xz) \ge 0$$ which is obviously true by Schur's ineq.
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Re: Inequality with a,b,c sides of a triangle
Sorry Atonu, but I don't understand some passages...
Why exactly
$$\sum_{cyc} \frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}}=
\sum_{cyc} \sqrt{1 \frac{(xy)(xz)}{2x^2}} \le \sum_{cyc} 1\frac{(xy)(xz)}{4x^2} = 3  \frac{1}{4} \sum_{cyc} x^{2}(xy)(xz) $$ ?
And how do you use exactly Shur's inequality to show that $$\sum_{cyc} x^{2}(xy)(xz) \ge 0$$ ?
Why exactly
$$\sum_{cyc} \frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}}=
\sum_{cyc} \sqrt{1 \frac{(xy)(xz)}{2x^2}} \le \sum_{cyc} 1\frac{(xy)(xz)}{4x^2} = 3  \frac{1}{4} \sum_{cyc} x^{2}(xy)(xz) $$ ?
And how do you use exactly Shur's inequality to show that $$\sum_{cyc} x^{2}(xy)(xz) \ge 0$$ ?
 Atonu Roy Chowdhury
 Posts: 64
 Joined: Fri Aug 05, 2016 7:57 pm
 Location: Chittagong, Bangladesh
Re: Inequality with a,b,c sides of a triangle
$ \sqrt{b}+\sqrt{c}\sqrt{a}=x \Rightarrow b+ca=x^2 2a2\sqrt{bc}+2\sqrt{ca}+2\sqrt{ab}= x^2  \frac{1}{2}(xy)(xz)$Katy729 wrote: ↑Fri Apr 20, 2018 10:24 pmSorry Atonu, but I don't understand some passages...
Why exactly
\[ \sum_{cyc} \frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}}=
\sum_{cyc} \sqrt{1 \frac{(xy)(xz)}{2x^2}} \le \sum_{cyc} (1\frac{(xy)(xz)}{4x^2}) = 3  \frac{1}{4} \sum_{cyc} x^{2}(xy)(xz) \] ?
And how do you use exactly Shur's inequality to show that $$\sum_{cyc} x^{2}(xy)(xz) \ge 0$$ ?
$\therefore \frac{\sqrt{b+ca}}{\sqrt{b}+\sqrt{c}\sqrt{a}} = \sqrt{\frac{x^2  \frac{1}{2}(xy)(xz)}{x^2}}=\sqrt{1  \frac{(xy)(xz)}{2x^2}}$
Let $\frac{(xy)(xz)}{x^2}=n$. Now, all I wanted to do is flip the root. That is, $\sqrt{1\frac{n}{2}}=\sqrt{12\frac{n}{4}} \le \sqrt{12\frac{n}{4}+\frac{n^2}{16}} = 1 \frac{n}{4}$.
And, Schur's ineq states that \[ \sum_{cyc}x^n(xy)(xz) \ge 0 \] for positive real $x,y,z$ and integer $n$.
This was freedom. Losing all hope was freedom.
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
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: 64
 Joined: Fri Aug 05, 2016 7:57 pm
 Location: Chittagong, Bangladesh
Re: Inequality with a,b,c sides of a triangle
Okay, the exponent must not be positive. One can easily prove that if the exponent is negative, the inequality still holds.Katy729 wrote: ↑Sat Apr 21, 2018 10:26 pmThanks 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
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.
Re: Inequality with a,b,c sides of a triangle
Yes, now understand. Very gentle Atonu!
 Atonu Roy Chowdhury
 Posts: 64
 Joined: Fri Aug 05, 2016 7:57 pm
 Location: Chittagong, Bangladesh
Re: Inequality with a,b,c sides of a triangle
Okay, you can check this out if you have confusion about the proof of negative exponent.
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This was freedom. Losing all hope was freedom.