Expand: \[\sum \frac{a^{2}}{b^{2}}+2\sum \frac{a}{b}\geq 3+\sum \frac{a}{b}+\sum \frac{b}{a}\iff \sum \frac{a^{2}}{b^{2}}+\sum \frac{a-c}{b}\geq 3 \] No time to post rest right now but I think I can prove the stronger one \[\sum \frac{(a+b)(a-c)}{b^{2}}\geq 0\] which directly implies this one...

Anything complete or better than just multiplying out?

Iran 2005

Wow, this is pretty easy for a MOP problem...

Karth wrote: $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}$ $\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 9$ This step is wrong because it's irreversible

Wouldn't it be reversible since $a/b+b/a \ge 2$?

Karth wrote: Wouldn't it be reversible since $a/b+b/a \ge 2$? You would have to prove $\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge 9 \ge 3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}$ which means you would have to prove $2\ge \frac{a}{b}+\frac{b}{a}$

In other words, you have conveniently weakened the bound that you are trying to prove. If we want to prove $a \geq b$ and we know $b \geq c$ and $a \geq c$, well, that does not solve our original problem.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=512906#p512906

Multiply by $(abc)^2$, so that we have $$(a^2b+b^2c+c^2a)^2\geq abc(a+b+c)(ab+ac+bc).$$Because I am not smart, we expand: $$a^4b^2+b^4c^2+c^4a^2+2(a^2b^3c+b^2c^3a+c^2a^3b)\geq (\sum_{sym}a^3b^2c) + 3a^2b^2c^2.$$Cancelling as many terms as possible, this becomes $$a^4b^2+b^4c^2+c^4a^2+a^2b^3c+b^2c^3a+c^2a^3b\geq a^3b^2c+b^3c^2a+c^3a^2b+3a^2b^2c^2\text{~~~~~(A)}$$Note that by AM-GM, we have $$\frac{2}{3}(a^4b^2)+\frac{1}{3}ab^2c^3\geq a^3b^2c,$$$$\frac{1}{3}(a^4b^2)+\frac{2}{3}ab^2c^3\geq a^2b^2c^2.$$Adding these two together, $$a^4b^2+ab^2c^3\geq a^3b^2c+a^2b^2c^2 \text{~~~~~(C)}.$$Then, (A) is just the sum of the cyclic variants of (C), so we are done.