# Inequality for $a,b,c>0$ $\sum_{cyc}\sqrt{\frac{a^3}{14a^2+4b^2}}\leq \sum_{cyc}\sqrt{\frac{a+b}{36}}$

A friend gives me the following result :

Let $$a,b,c>0$$ then we have : $$\sqrt{\frac{a^3}{14a^2+4b^2}}+\sqrt{\frac{b^3}{14b^2+4c^2}}+\sqrt{\frac{c^3}{14c^2+4a^2}}\leq \sqrt{\frac{a+b}{36}}+\sqrt{\frac{b+c}{36}}+\sqrt{\frac{c+a}{36}}$$

He tells me also that we can use majorization to prove this (with $$f(x)=-\sqrt{x}$$) furthermore I can prove the last line of the majorization which is : $$\frac{a^3}{14a^2+4b^2}+\frac{b^3}{14b^2+4c^2}+\frac{c^3}{14c^2+4a^2}\geq \frac{a+b}{36}+\frac{b+c}{36}+\frac{c+a}{36}$$ But i'm stuck with the two first lines of the majorization .

An other way is to study an inequality with two variables (if we divide the two sides by $$\sqrt{a}$$) and then we fix a variable to get a one variable inequality and to get the minimum we study the derivative .

It's all my ideas but in fact I would like an Olympiad proof or something like that (if you can obviously) .

Thanks.

Edit : It's not $$39$$ but $$36$$ obviously...

• If a=b=c=1,your first inequality is wrong. – Bonbon May 9 at 14:06
• $\sum\limits_{cyc}\frac{a^3}{14a^2+4b^2}\geq\sum\limits_{cyc}\frac{a+b}{37}$ is also true. – Michael Rozenberg May 9 at 18:52
• @MichaelRozenberg, I think above is from: $\lceil$ math.stackexchange.com/questions/1777075/… $\rfloor$, right? – user685500 May 10 at 10:32
• @max8128 Now your second inequality is wrong. – Michael Rozenberg May 10 at 13:22
• This is typical FatsWallers - a random inequality, wrong, which keeps changing based on feedback. – Macavity May 12 at 16:19

## 1 Answer

We need to prove that: $$3\sqrt2\sum_{cyc}\sqrt{\frac{a^3}{7a^2+2b^2}}\leq\sum_{cyc}\sqrt{a+b}.$$ Now, by C-S $$\sqrt{\frac{a^3}{7a^2+2b^2}}\leq\sqrt{\sum_{cyc}(7a^2+2c^2)\sum_{cyc}\frac{a^3}{(7a^2+2b^2)(7a^2+2c^2)}}=3\sqrt{\sum_{cyc}a^2\sum_{cyc}\frac{a^3}{(7a^2+2b^2)(7a^2+2c^2)}}$$ and by C-S again and by AM-GM $$\sum_{cyc}\sqrt{a+b}=\sqrt{\sum_{cyc}(a+b+2\sqrt{(a+b)(a+c)}}=$$ $$=\sqrt{2\sum_{cyc}a+2\sqrt{\sum_{cyc}\left((a+b)(a+c)+2(a+b)\sqrt{(a+c)(b+c)}\right)}}\geq$$ $$\geq\sqrt{2\sum_{cyc}a+2\sqrt{\sum_{cyc}\left(a^2+3ab+2(a+b)(\sqrt{ab}+c)\right)}}=$$ $$=\sqrt{2\sum_{cyc}a+2\sqrt{\sum_{cyc}\left(a^2+7ab+2(\sqrt{a^3b}+\sqrt{ab^3})\right)}}\geq$$ $$\geq\sqrt{2\sum_{cyc}a+2\sqrt{\sum_{cyc}\left(a^2+7ab+2(2ab)\right)}}=\sqrt{2\sum_{cyc}a+2\sqrt{\sum_{cyc}\left(a^2+11ab\right)}}.$$ Thus, it's enough to prove that $$81(a^2+b^2+c^2)\sum_{cyc}\frac{a^3}{(7a^2+2b^2)(7a^2+2c^2)}\leq a+b+c+\sqrt{\sum_{cyc}(a^2+11ab)}.$$ Now, we'll prove that $$\sqrt{\sum_{cyc}(a^2+11ab)}\geq \frac{(a+b+c)\sum\limits_{cyc}(a^2+7ab)}{\sum\limits_{cyc}(a^2+3ab)}.$$ Indeed, let $$a^2+b^2+c^2=k(ab+ac+bc).$$

Thus, we need to prove here that $$\sqrt{k+11}\geq \frac{\sqrt{k+2}(k+7)}{k+3}$$ or $$(k+11)(k+3)^2\geq(k+2)(k+7)^2$$ or $$(k-1)^2\geq0.$$ Id est, it's enough to prove that $$a+b+c+ \frac{(a+b+c)\sum\limits_{cyc}(a^2+7ab)}{\sum\limits_{cyc}(a^2+3ab)}\geq81(a^2+b^2+c^2)\sum_{cyc}\frac{a^3}{(7a^2+2b^2)(7a^2+2c^2)},$$ which is true by BW.

Indeed, let $$a=\min\{a,b,c\}$$, $$b=a+u$$ and $$c=a+v$$.

After this substitution we obtain something obvious:

https://www.wolframalpha.com/input/?i=(x%2By%2Bz)(2(x%5E2%2By%5E2%2Bz%5E2)%2B10(xy%2Bxz%2Byz))-81(x%5E2%2By%5E2%2Bz%5E2%2B3(xy%2Bxz%2Byz))(x%5E2%2By%5E2%2Bz%5E2)(x%5E3%2F((7x%5E2%2B2y%5E2)(7x%5E2%2B2z%5E2))%2By%5E3%2F((7y%5E2%2B2z%5E2)(7y%5E2%2B2x%5E2))%2Bz%5E3%2F((7z%5E2%2B2x%5E2)(7z%5E2%2B2y%5E2))),x%3Da,y%3Da%2Bu,z%3Da%2Bv

Your second inequality is wrong.

Try $$(a,b,c)=(4,1,2).$$

By the way, the following inequality is true already and it's not so trivial.

For positives $$a$$, $$b$$ and $$c$$ prove that: $$\frac{a^3}{14a^2+4b^2}+\frac{b^3}{14b^2+4c^2}+\frac{c^3}{14c^2+4a^2}\geq\frac{a+b}{37}+\frac{a+c}{37}+\frac{b+c}{37}.$$