Prove $\sqrt{\frac{a+3}{a+3b}}+\sqrt{\frac{b+3}{b+3c}}+\sqrt{\frac{c+3}{c+3a}} \ge 3$

Question

Let $a,b,c$ are positives such that $ab+bc+ca=3$. Prove that: $$\sqrt{\frac{a+3}{a+3b}}+\sqrt{\frac{b+3}{b+3c}}+\sqrt{\frac{c+3}{c+3a}} \ge 3$$ Once, I see this problem in AoPS:https://artofproblemsolving.com/community/c6h3248614p29942459

I thought it was easy until I decide to solve it. Here is some of my attempt:

First attempt: By using AM-GM, we have: $$\sqrt{\frac{a+3}{a+3b}}+\sqrt{\frac{b+3}{b+3c}}+\sqrt{\frac{c+3}{c+3a}} \ge 3\sqrt[6]{\frac{(a+3)(b+3)(c+3)}{(a+3b)(b+3c)(c+3a)}}$$ So we just need to prove: $$(a+3)(b+3)(c+3) \ge (a+3b)(b+3c)(c+3a)$$ Which is obviously wrong.

Second attempt: By Bernoulli inequality, we have: $$\sqrt{\frac{a+3}{a+3b}} \ge 1+\frac{3(b-1)}{2(a+3b)}$$ So the first inequality is equivalent to: $$\sum_{cyc} \frac{3(b-1)}{2(a+3b)} \leq 0$$ which is equivalent to: $$\sum_{cyc }a^2 b + 6 \sum_{cyc}a^2 c+27 a b c \leq 39+3\sum_{cyc}a^2$$ Fortunately, it's obvious, but there is a big mistake here

This is Bernoulli inequality is reversed for $h \le1$, so my second attempt is failed.

Additionally, it's also a proof using Holder by sir arqady, but I and the author want to find a nicer proof for this inequality.

Thank you very much!

Answer 1

Some thoughts.

By Holder inequality, we have $$\left(\sum_{\mathrm{cyc}} \sqrt{\frac{a+3}{a+3b}}\right)^2 \sum_{\mathrm{cyc}} (a + 3b)(a + 3)^2(c + 2)^3 \ge \left(\sum_{\mathrm{cyc}} (a + 3)(c + 2)\right)^3. $$

It suffices to prove that $$\left(\sum_{\mathrm{cyc}} (a + 3)(c + 2)\right)^3 \ge 9 \sum_{\mathrm{cyc}} (a + 3b)(a + 3)^2(c + 2)^3.$$ This inequality is true which is verified by Mathematica (a Computer Algebra System). A human verfiable proof is required.