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

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

1 Answer 1


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?

  • $\begingroup$ This was no modest one, but you defeated easily enough. What style ! $\endgroup$
    – user688846
    Jul 15, 2019 at 3:43
  • $\begingroup$ Could you tell more about the last step AM-GM ? $\endgroup$ Jul 17 at 9:32
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    $\begingroup$ Thank you, it is equivalent to $(w-1)^2.f(w)\ge 0$ $\endgroup$ Jul 17 at 10:49
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    $\begingroup$ 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. $\endgroup$ Jul 17 at 13:34
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    $\begingroup$ Thank you. I saw you used Holder a beautiful and I tried. $\endgroup$ Jul 17 at 13:40

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