How to prove $a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+\sqrt[3]{abc}\ge 7\sqrt[7]{\frac{abc(a+b)(b+c)(c+a)(a+b+c)}{24}}$?

Question

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.

Answer 1

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 $a=b=c.$

Answer 2

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.