2

If $a,b,c\ge 0,$ then prove $$a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+\sqrt[3]{abc}\ge 7\sqrt[7]{\frac{abc(a+b)(b+c)(c+a)(a+b+c)}{24}}$$ I found a problem here and try to use AM-GM but it didn't work.

By exponential, it leads to very complicated things.

Is there nice proof like AM-GM, ...?

All comment and answer is welcome. Thanks.

Michael Rozenberg
  • 194,933
Dragon boy
  • 1
  • 16

2 Answers2

1

The first step may be the following AM-GM. $$LHS\geq7\sqrt[7]{\left(\frac{a+b+c}{3}\right)^3\left(\frac{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}}{3}\right)^3\sqrt[3]{abc}}$$ and it's enough to prove that: $$8(a+b+c)^2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})^3\geq243(a+b)(a+c)(b+c)\sqrt[3]{a^2b^2c^2},$$ but my proof of this statement is not nice.

Michael Rozenberg
  • 194,933
  • Did you check the last inequality ? – Dragon boy Aug 10 '23 at 05:14
  • Yes, it's true. I proved it and I am looking for a nice proof. – Michael Rozenberg Aug 10 '23 at 05:25
  • @Dragonboy Michael Rozenberg's last inequality is right,because it is equivalent to $$ \frac{1}{386}\sum (a-b)^{2}(3088a^6b^3+3396a^5b^4+25578a^5b^3c+3396a^4b^4c+3088a^3b^6+25578a^3b^5c+5196ac^6(a-b)^2+41b^2c(c-a)^4(a-b)^2+5196bc^6(a-b)^2+41bc^2(c-a)^4(a-b)^2+708c^7(a-b)^2) +\frac{1}{386}\sum a(a^2-bc)(6484a^3b^2c+6484a^3bc^2+1088ab(b-c)^4+1088ac(b-c)^4+159(b-c)^6)+\frac{1290}{193}\sum a(a^3-bc^2)^2(b-c)^4+\frac{1290}{193}\sum a(a^3-b^2c)^2(b-c)^4\ge 0 $$ – Pesilk Aug 10 '23 at 05:30
  • @MichaelRozenberg Did you prove it by $uvw$ ? – Dragon boy Aug 10 '23 at 05:38
  • @Pesilk How may I learn about program from you ? – Dragon boy Aug 10 '23 at 05:38
0

Proof.

Firstly, we'll prove $$\left[(\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{b})(\sqrt{a}+\sqrt{c})\right]^4\ge 512abc(a+b)(b+c)(c+a).\tag{*}$$ Indeed, by AM-GM $$(\sqrt{a}+\sqrt{b})^4=(a+b+2\sqrt{ab})^2\ge 8\sqrt{ab}(a+b).$$Multiplying similar inequalities, the $(*)$ is proven.

Now, apply $(*)$ and AM-GM as \begin{align*} 7\sqrt[7]{\frac{abc(a+b)(b+c)(c+a)(a+b+c)}{24}}=&7\left[\sqrt[3]{abc}.\left(\sqrt[6]{\frac{abc(a+b)(b+c)(c+a)}{8}}\right)^4.\frac{a+b+c}{3}.\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}\right]^{\frac{1}{7}}\\&\le \sqrt[3]{abc}+\sqrt[3]{(\sqrt{a}+\sqrt{b})^2(\sqrt{c}+\sqrt{b})^2(\sqrt{a}+\sqrt{c})^2}+\frac{a+b+c}{3}+\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}\\&\le \frac{a+b+c}{3}+\frac{a+b+c}{3}+\sqrt[3]{abc}+\frac{\sum\limits_{cyc}(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}{3}\\&=a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+\sqrt[3]{abc}. \end{align*} Hence, the proof is done. Equality holds at $a=b=c.$

TATA box
  • 1
  • 1
  • 5
  • 29