$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3}{\sqrt{2}}.$ explaination

Question

Let $a,b,c$ are non-negative numbers such that: $a+b+c=3$. Prove that: $$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3}{\sqrt{2}}.$$

Solution:

Using Cauchy-Schwarz inequality:

$$\left(\sum _{cyc}\sqrt{\frac{a}{a+b}} \right)^2 =\left(\sum _{cyc}\sqrt{(c+a).\frac{a}{(a+b)(a+c)}}\right)^2\leq \left[\sum_{cyc}(c+a)\right]\left[\sum_{cyc}\frac{a}{(a+b)(a+c)}\right]$$ $$=\frac{4\left(\sum_{cyc}a\right)\left(\sum_{cyc}ab\right)}{(a+b)(b+c)(c+a)}=\frac{4\left(\sum_{cyc}a\right)\left(\sum_{cyc}ab\right)}{\left(\sum_{cyc}a\right)\left(\sum_{cyc}ab\right)-abc}\leq \frac{4\left(\sum_{cyc}a\right)\left(\sum_{cyc}ab\right)}{\frac{8}{9}\left(\sum_{cyc}a\right)\left(\sum_{cyc}ab\right)}=\frac{9}{2}.$$

So we have q.e.d.

But my question is how can they find this:$$\left(\sum _{cyc}\sqrt{\frac{a}{a+b}} \right)^2 =\left(\sum _{cyc}\sqrt{(c+a).\frac{a}{(a+b)(a+c)}}\right)^2\leq \left[\sum_{cyc}(c+a)\right]\left[\sum_{cyc}\frac{a}{a+b}\right]$$

By using:$$\sqrt{\frac{a}{a+b}}=\sqrt{(a+c).\frac{a}{(a+b)(a+c)}}$$

How can they know that?

It seems that this is your first time using internet so let me tell you that writing in ALLCAPS in internet is the equivalent of yelling. Please, don't yell at us. — jjagmath, Oct 27 '23 at 20:12

score 2 · Accepted Answer · answered Oct 27 '23 at 16:53

2

Because $\sum\limits_{cyc}\frac{a}{a+b}$ is cyclic and if we want to get something symmetric so the first idea is to try to work with $\sum\limits_{cyc}\frac{a}{(a+b)(a+c)},$ which gives your C-S.

In the general there is a Bacteria's method which sometimes helps if the equality occurs in another points (or closed to another point) not only for $a=b=c$.

In your inequality it's not so and it's enough to think about symmetrization.

answered Oct 27 '23 at 16:53

Michael Rozenberg

194,933

But how can the find the $(a+c)$ and why don't they find an another things like $a^2+bc, a^2+b^2+c^2,...$. Thank you a lots! – Lục Trường Phát Oct 28 '23 at 15:01
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@Lục Trường Phát $\sum\limits_{cyc}\frac{a}{(*a+b)(a+c)}$ is symmetric. See please better my post. With this trick there are another solutions, of course, but they are not so nice. There is a nice solution by Jensen. – Michael Rozenberg Oct 28 '23 at 15:47
Where can I find the Bacteria's method? Thank a lots!!! – Lục Trường Phát Oct 29 '23 at 07:18
1

About Bacteria see here: https://math.stackexchange.com/questions/2903914/what-are-the-most-popular-techniques-of-proving-inequalities – Michael Rozenberg Oct 29 '23 at 09:03

score 2 · Answer 2 · answered Oct 27 '23 at 19:59

2

That's because it's the wrong statement. The application of CS is actually

$$\left(\sum _{cyc}\sqrt{(c+a).\frac{a}{(a+b)(a+c)}}\right)^2\leq \left[\sum_{cyc}(c+a)\right]\left[\sum_{cyc}\frac{a}{(a+b)\color{red}{(a+c)}}\right]$$

To get to the next line, observe that $\sum (c+a) = 2 \sum a$ and $\sum \frac{a}{(a+b)(a+c) } = \sum \frac{ ab+ac}{(a+b)(b+c)(c+a) } = \frac{ 2(\sum ab)}{(a+b)(b+c)(c+a)}$. Thus

$$\left[\sum_{cyc}(c+a)\right]\left[\sum_{cyc}\frac{a}{(a+b)\color{red}{(a+c)}}\right]= \frac{4\left(\sum_{cyc}a\right)\left(\sum_{cyc}ab\right)}{\left(\sum_{cyc}a\right)\left(\sum_{cyc}ab\right)-abc}$$

annd the rest of the proof is fine.

answered Oct 27 '23 at 19:59

Calvin Lin

68,864

Actually that is my mistake in writting, can you explain it please? Thank you! – Lục Trường Phát Oct 28 '23 at 15:00
1

@LụcTrườngPhát Ah, if that was a typo on your end, then Michael's post dives into that. There's much to learn from his experience of solving inequalities. – Calvin Lin Oct 30 '23 at 18:50

$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3}{\sqrt{2}}.$ explaination

2 Answers2