0
$\begingroup$

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...

$\endgroup$
  • 4
    $\begingroup$ 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
  • 4
    $\begingroup$ This is typical FatsWallers - a random inequality, wrong, which keeps changing based on feedback. $\endgroup$ – Macavity May 12 at 16:19
3
+50
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.