2
$\begingroup$

prove $$\sum_{cyc}\frac{a^2}{a+2b^2}\ge 1$$ holds for all positives $a,b,c$ when $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ or $ab+bc+ca=3$


Background Taking $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ This was left as an exercise to the reader in the book 'Secrets in inequalities'.This comes under the section Cauchy Reverse technique'.i.e the sum is rewritten as :$$\sum_{cyc} a- \frac{ab^2}{a+2b^2}\ge a-\frac{2}{3}\sum_{cyc}{(ab)}^{2/3}$$ which is true by AM-GM.($a+b^2+b^2\ge 3{(a. b^4)}^{1/3}$)

By QM-AM inequality $$\sum_{cyc}a\ge \frac{{ \left(\sum \sqrt{a} \right)}^2}{3}=3$$.

we are left to prove that $$\sum_{cyc}{(ab)}^{2/3}\le 3$$ .But i am not able to prove this .Even the case when $ab+bc+ca=3$ seems difficult to me.

Please note I am looking for a solution using this cuchy reverse technique and AM-GM only.

$\endgroup$
3
  • $\begingroup$ I think you can prove by AM GM that $$\dfrac{1}{3}\sum x^{2/3} \le\left(\dfrac{1}{3} \sum x\right)^{2/3}$$ by expanding with the third power and replace $x=ab$ $\endgroup$ Oct 6, 2020 at 9:03
  • $\begingroup$ @BookOfFlames Botha are differnt problems $\endgroup$ Oct 6, 2020 at 9:14
  • $\begingroup$ Yes, I did not see it. $\endgroup$ Oct 6, 2020 at 9:16

2 Answers 2

3
$\begingroup$

Now, for $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ we need to prove that: $$\sum_{cyc}\sqrt[3]{a^2b^2}\leq3$$ or $$\sum_{cyc}\sqrt[3]{a^4b^4}\leq3,$$ where $a$, $b$ and $c$ are positives such that $a+b+c=3,$ which is true by AM-GM and Schur: $$\sum_{cyc}\sqrt[3]{a^4b^4}=\sum_{cyc}ab\sqrt[3]{ab}\leq\frac{1}{3}\sum_{cyc}ab(1+a+b)\leq3.$$

$\endgroup$
2
  • $\begingroup$ thank you for answer,but is there a way to eliminate schur? $\endgroup$ Oct 6, 2020 at 9:13
  • $\begingroup$ @Albus Dumbledore Instead of Schur we can use SOS, uvw, BW and more...But I don't see a solution for the last step by AM-GM. $\endgroup$ Oct 6, 2020 at 9:15
3
$\begingroup$

For $ab+bc+ca=3.$ We need to prove $$ a+b+c \geqslant 1 + \frac{2}{3}\sum (ab)^{2/3}.$$ From condition we get $$a+b+c \geqslant \sqrt{3(ab+bc+ca)}=3.$$ Now, using the AM-GM inequality we have $$\sum (ab)^{2/3} = \sum \sqrt[3]{1 \cdot ab \cdot ab} \leqslant \sum \frac{1+2ab}{3} = 1+\frac{2}{3}(ab+bc+ca)=3.$$ So $$1 + \frac{2}{3}\sum (ab)^{2/3} \leqslant 1 + \frac{2}{3} \cdot 3=3 \leqslant a+b+c.$$ Done.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .