Inequality about $\sum_{cyc}\dfrac{a^3}{(bc)^2+1}$ and $a+b+c=ab+bc+ca.$

Question

Context.

My friend sent me problem which hard for me to think of good approach. It's

Let $a,b,c>0:a+b+c=ab+bc+ca.$ Prove that$$\frac{a^3}{(bc)^2+1}+\frac{b^3}{(ca)^2+1}+\frac{c^3}{(ab)^2+1}+3\ge \frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{2}.$$ Here's what I've done so far.

By using AM-GM $$\frac{2a^3}{(bc)^2+1}+\frac{(bc)^2+1}{2}+1\ge 3a.$$ It implies$$\frac{a^3}{(bc)^2+1}\ge \frac{3a}{2}-\frac{(bc)^2}{4}-\frac{3}{4}.$$ Sum up analogs, we obtain$$\frac{a^3}{(bc)^2+1}+\frac{b^3}{(ca)^2+1}+\frac{c^3}{(ab)^2+1}+3\ge \frac{3(a+b+c)}{2}-\frac{(ab)^2+(bc)^2+(ca)^2}{4}+\frac{3}{4}.$$ I got stuck to prove $$\frac{3(a+b+c)}{2}-\frac{(ab)^2+(bc)^2+(ca)^2}{4}+\frac{3}{4}\ge \frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{2}.$$ It's $$\frac{a+b+c}{2}+\frac{3}{4}\ge \frac{(ab)^2+(bc)^2+(ca)^2+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}}{4}.$$ Could you please help me prove the last inequality?

Also, in my idea I don't see how to use the condition $a+b+c=ab+bc+ca.$

Hope to see some better ideas. Thank you.

Answer 1

Proof.

By using AM-GM $$\frac{2a^3}{(bc)^2+1}+\sqrt{\frac{(bc)^2+1}{2}}+\sqrt{\frac{(bc)^2+1}{2}}\ge 3a.$$ It implies$$\frac{a^3}{(bc)^2+1}\ge \frac{3a}{2}-\frac{\sqrt{2}}{2}\sqrt{(bc)^2+1}.\tag{1}$$ Now, we may use AM-GM again.

Notice that$$\sqrt{(bc)^2+1}=\sqrt{(bc+1)^2-(\sqrt{2bc})^2}=\sqrt{(bc+1-\sqrt{2bc})\cdot(bc+1+\sqrt{2bc})}$$ $$=\sqrt{2}\cdot\sqrt{\frac{bc+1-\sqrt{2bc}}{2-\sqrt{2}}\cdot\frac{bc+1+\sqrt{2bc}}{2+\sqrt{2}}}\le \frac{\sqrt{2}}{2}\left(\frac{bc+1-\sqrt{2bc}}{2-\sqrt{2}}+\frac{bc+1+\sqrt{2bc}}{2+\sqrt{2}}\right)$$ $$=\frac{\sqrt{2}}{2}\left(2+2bc-2\sqrt{bc}\right)=\sqrt{2}\cdot\left(1+bc-\sqrt{bc}\right). \tag{2}$$ From $(1)$ and $(2),$ we get $$\frac{a^3}{(bc)^2+1}\ge \frac{3a}{2}-1-bc+\sqrt{bc}.\tag{*}$$ Take cyclic sum on $(*)$ $$\frac{a^3}{(bc)^2+1}+\frac{b^3}{(ca)^2+1}+\frac{c^3}{(ab)^2+1}+3\ge \frac{3(a+b+c)-2(ab+bc+ca)}{2}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca},$$ which the given condition gives proof here. Equality holds iff $a=b=c=1.$

Answer 2

The inequality, which you got is wrong.

Try $c=0$ and $a=b=2$.

About the starting problem:

By C-S $$\sum_{cyc}\frac{a^3}{b^2c^2+1}=\sum_{cyc}\frac{a^4}{ab^2c^2+a}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2c+a)}.$$ Also, by C-S again $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2\leq3(a+b+c)$$ and it's enough to prove that: $$\frac{(a^2+b^2+c^2)^2}{(a+b+c)(abc+1)}+3\geq\frac{3(a+b+c)}{2}.$$ Now, let $a+b+c=3u$.

Thus, $u\geq1$ and by AM-GM $$abc\leq\frac{(a+b+c)(ab+ac+bc)}{9}=u^2$$ and it's enough to prove that: $$\frac{(9u^2-6u)^2}{3u(u^2+1)}+3\geq\frac{9u}{2}$$ or $$(u-1)(3u-1)(5u+1)\geq0$$ and we are done.