Let $a,b,c\ge 0: a+b+c=3.$ Prove that$$\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}.$$ This problem is from a book.

I tried to used AM-GM without any success.

Notice that $3(ab+bc+ca)-abc=(a+b)(b+c)(c+a).$ By AM-GM for three variables$$\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge 3\sqrt[6]{\prod_{cyc}(a+b).\prod_{cyc}(ab+2)}.$$We need to prove$$27\prod_{cyc}(a+b)\ge 8\prod_{cyc}(ab+2),$$which is wrong at $a=0;b=c=\dfrac{3}{2}.$

I hope to see better ideas. Thank you.


1 Answer 1


Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, by your work we need to prove that $$\sum_{cyc}\left(a^2b+a^2c+4+2\sqrt{(a+b)(a+c)(ab+2)(ac+2)}\right)\geq\frac{27}{4}\prod_{cyc}(a+b).$$ But by AM-GM $$2\sum_{cyc}\sqrt{(a+b)(a+c)(ab+2)(ac+2)}\geq6\sqrt[3]{\prod_{cyc}((a+b)(ab+2))}$$ and it's enough to prove that $$\sum_{cyc}(a^2b+a^2c+4a)+6\sqrt[3]{\prod_{cyc}((a+b)(ab+2))}\geq\frac{27}{4}\prod_{cyc}(a+b)$$ or $$9uv^2-3w^3+12u^3+6\sqrt[3]{(9uv^2-w^3)\prod_{cyc}(ab+2u^2)}\geq\frac{27}{4}(9uv^2-w^3)$$ or $$4(4u^3+3uv^2-w^3)-9(9uv^2-w^3)+8\sqrt[3]{(9uv^2-w^3)(w^6+3uw^3\cdot2u^2+3v^2\cdot4u^4+8u^6)}\geq0$$ or $f(w^3)\geq0,$ where $$f(w^3)=(16u^3-69uv^2+5w^3)^3+512(9uv^2-w^3)(w^6+6u^3w^3+12u^4v^2+8u^6).$$ But $$f''(w^3)=-3744u^3-1134uv^2-2322w^3<0,$$ which says that $f$ is a concave function and from here $f$ gets a minimal value for an extremal value of $w^3$, which by $uvw$ says that it's enough to prove $f(w^3)\geq0$ in the following cases.

  1. $w^3=0$.

Let $c=0$.

Thus, $b=3-a$, where $0\leq a\leq3$ and we need to prove that $$a^2(3-a)+(3-a)^2a+12+6\sqrt[3]{3a(3-a)(a(3-a)+2)\cdot4}\geq\frac{27}{4}\cdot3a(3-a),$$ which is true even for any real value of $a$.

  1. Two variables are equal.

Let $b=c=1$ after homogenization.

Thus, it's enough to prove that: $$\left(\frac{16(a+2)^3}{27}-\frac{23(a+2)(2a+1)}{3}+5a\right)^3+$$ $$+512((a+2)(2a+1)-a)\left(a^2+\frac{2(a+2)^3a}{9}+\frac{4(a+2)^4(2a+1)}{81}+\frac{8(a+2)^6}{729}\right)\geq0$$ or $$(a-1)^2(512a^7-1856a^6+1170456a^5+6520049a^4+13118644a^3+12182070a^2+5225276a+835921)\geq0,$$ which is obvious.

  • $\begingroup$ The hardest step is squaring both side and use AM-GM. I did not think of it. Thank you very much, @Michael Rozenberg. $\endgroup$ Jul 21 at 16:55
  • $\begingroup$ @Ha Diep Xuan You are welcome! $\endgroup$ Jul 21 at 16:55
  • $\begingroup$ Just a thought: We can write the OP as $$\sqrt{\frac{ab+2}{ab+3c}}+\sqrt{\frac{bc+2}{bc+3a}}+\sqrt{\frac{ca+2}{ca+3b}}\ge \frac{3\sqrt{3}}{2}.$$ Did you try Holder,@Michael Rozenberg ? $\endgroup$ Jul 22 at 7:43
  • $\begingroup$ Yes, I tried it. Holder does not help here. $\endgroup$ Jul 22 at 9:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .