If $ab+bc+ca=1,$ prove $\sum_{cyc}\frac{1+a\sqrt{bc}}{a+\sqrt{bc}}\ge 1+\frac{4}{a+b+c}.$

Question

Problem. Prove that the following inequality$$\frac{1+a\sqrt{bc}}{a+\sqrt{bc}}+\frac{1+b\sqrt{ca}}{b+\sqrt{ca}}+\frac{1+c\sqrt{ab}}{c+\sqrt{ab}}\ge 1+\frac{4}{a+b+c},$$holds for all non-negative real numbers $a,b,c$ such that $ab+bc+ca=1.$

---

It is Problem 4840, CRUX Mathematicorum.

I've tried to verify the original form as$$\sum_{cyc}\left(\frac{1+a\sqrt{bc}}{a+\sqrt{bc}}+1\right)\ge 4\left(1+\frac{1}{a+b+c}\right),$$$$\iff \sum_{cyc}\frac{(1+a)(1+\sqrt{bc})}{a+\sqrt{bc}}\ge 4\left(1+\frac{1}{a+b+c}\right).$$ I checked the inequality is true: $$\frac{1+\sqrt{bc}}{a+\sqrt{bc}}+\frac{1+\sqrt{ca}}{b+\sqrt{ca}}+\frac{1+\sqrt{ab}}{c+\sqrt{ab}}\ge 4.$$ It seems Chebyshev is appropriate $$\sum\limits_{cyc}(1+a);\sum\limits_{cyc}\frac{1+\sqrt{bc}}{a+\sqrt{bc}},$$but I am still stuck here.

Hope to see some ideas. Thank you.

Answer 1

I have read a technical of Khuong Trang called RAG (Root AM GM). Here what it says:

For $x,y,z \ge 0$ then we get $$\frac{x^2+y}{x+z} \ge 2\left(\sqrt{y+z^2}-z\right).$$ which can be proved by AM GM. Now set $$(x,y,z) \leftarrow (a,ab+ac,\sqrt{bc})$$ then $$\frac{a^2+ab+ac}{a+\sqrt{bc}} \ge 2\left(\sqrt{ab+bc+ca}-\sqrt{bc}\right),$$ Note that $$\sqrt{ab+bc+ca}-\sqrt{bc} =\frac{ab+ac}{\sqrt{ab+bc+ca}+\sqrt{bc}},$$ so $$\frac{a(a+b+c)}{a+\sqrt{bc}} \ge \frac{2a(b+c)}{\sqrt{ab+bc+ca}+\sqrt{bc}},$$ or $$\frac{(a+b+c)}{a+\sqrt{bc}} \ge \frac{2(b+c)}{\sqrt{ab+bc+ca}+\sqrt{bc}},$$ or $$\frac{\sqrt{ab+bc+ca}+\sqrt{bc}}{a+\sqrt{bc}}\ge \frac{2(b+c)}{a+b+c}$$

By the way, I think this is the way how the author created this problem.

Answer 2

By using your work we'll prove that $$\frac{1+\sqrt{bc}}{a+\sqrt{bc}}\geq\frac{2(b+c)}{a+b+c}.$$ Indeed, we need to prove that $$(a+b+c)\sqrt{ab+ac+bc}\geq2(b+c)(a+\sqrt{bc})-\sqrt{bc}(a+b+c)$$ or $$(a+b+c)\sqrt{ab+ac+bc}\geq a(2b-\sqrt{bc}+2c)+\sqrt{bc}(b+c),$$ Which is true by C-S: $$(a+b+c)\sqrt{ab+ac+bc}=\sqrt{(4ab+4ac+(b+c-a)^2)(ab+ac+bc)}\geq$$ $$\geq2(ab+ac)+\sqrt{bc}(b+c-a)=a(2b-\sqrt{bc}+2c)+\sqrt{bc}(b+c).$$ From here we obtain: $$\frac{(1+\sqrt{bc})(1+a)}{a+\sqrt{bc}}\geq\frac{2(b+c)(1+a)}{a+b+c}$$ or $$\frac{(1+\sqrt{bc})(1+a)}{a+\sqrt{bc}}-1\geq\frac{2(b+c)(1+a)}{a+b+c}-1$$ or $$\frac{1+a\sqrt{bc}}{a+\sqrt{bc}}\geq\frac{b+c-a+2(ab+ac)}{a+b+c},$$ which says: $$\sum_{cyc}\frac{1+a\sqrt{bc}}{a+\sqrt{bc}}\geq\sum_{cyc}\frac{b+c-a+2(ab+ac)}{a+b+c}=1+\frac{4}{a+b+c}.$$