Given three positive numbers $a,b,c$. Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{b+1}}\geqq3\sqrt[3]{\frac{4\,abc}{3\,abc+1}}$ .

Ji Chen. Given three positive numbers $$a, b, c$$. Prove that $$\sum\limits_{cyc}\sqrt{\frac{a+ b}{b+ 1}}\geqq 3\sqrt[3]{\frac{4\,abc}{3\,abc+ 1}}$$

Of course, we've to solve it by $$uvw$$, before that, I tried to use Holder-inequality with integer polynomials but without a high probability of success for me against this particular problem ...

I found here: https://artofproblemsolving.com/community/c6h538065p3209975, something obvious

• What is $uvw$ ? Commented Jul 15, 2019 at 3:17
• $uvw$ is a very useful method for the proof of polynomial inequalities with three variables. Sometimes it works for more variables as well. . .
– user688846
Commented Jul 15, 2019 at 3:18

By AM-GM $$\sum_{cyc}\sqrt{\frac{a+b}{b+1}}\geq3\sqrt[6]{\prod\limits_{cyc}\frac{a+b}{a+1}}.$$ Thus, it's enough to prove that $$(a+b)(a+c)(b+c)(3abc+1)^2\geq16a^2b^2c^2(a+1)(b+1)(c+1).$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.

Thus, we need to prove that $$(9uv^2-w^3)(3w^3+1)^2\geq16w^6(w^3+3v^2+3u+1)$$ and since by AM-GM $$uv^2\geq w^3,$$ it's enough to prove that $$uv^2(3w^3+1)^2\geq2w^6(w^3+3v^2+3u+1),$$ which is true by AM-GM.

Indeed, by AM-GM $$2w^6(w^3+3v^2+3u+1)=2w^9+6v^2w^6+6uw^6+2w^6\leq$$ $$\leq2uv^2w^6+6uv^2w^5+6uv^2w^4+2uv^2w^3$$ and it's enough to prove that: $$(3w^3+1)^2\geq2w^6+6w^5+6w^4+2w^3$$ or $$(3w^3+1)^2\geq2w^3(w+1)^3.$$ Can you end it now?

• This was no modest one, but you defeated easily enough. What style !
– user688846
Commented Jul 15, 2019 at 3:43
• Could you tell more about the last step AM-GM ? Commented Jul 17, 2023 at 9:32
• Thank you, it is equivalent to $(w-1)^2.f(w)\ge 0$ Commented Jul 17, 2023 at 10:49
• It's $\sum\limits_{sym}(a^7b+3a^6b^2-3a^4b^4+2a^6bc+3a^5b^2c-3a^4b^3c-3a^3b^3c^2)\geq0,$ which is obvious by Muirhead. Commented Jul 17, 2023 at 13:34
• Thank you. I saw you used Holder a beautiful and I tried. Commented Jul 17, 2023 at 13:40