# prove $\sum_{cyc}\frac{a^2}{a+2b^2}\ge 1$

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.

• 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 $x=ab$ Oct 6, 2020 at 9:03
• @BookOfFlames Botha are differnt problems Oct 6, 2020 at 9:14
• Yes, I did not see it. Oct 6, 2020 at 9:16

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