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) .


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

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

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:


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: $$\frac{a^3}{14a^2+4b^2}+\frac{b^3}{14b^2+4c^2}+\frac{c^3}{14c^2+4a^2}\geq\frac{a+b}{37}+\frac{a+c}{37}+\frac{b+c}{37}.$$


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