$a,b,c >0$, and $ab+bc+ca=3$, prove $(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$

Question

$a,b,c >0$, and $ab+bc+ca=3$, prove $$(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$

I think the equality is only achieve when $a=b=c=1$. The condition $ab+bc+ca=3$ is necessary. I used the estimation $x^x \geqslant \frac12 (x^2+1)$ but cannot proceed further.

Answer 1

Firstly, we'll prove that: $$\left(a^ab^bc^c\right)^{\frac{3}{a+b+c}}\geq\left(\frac{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}{3}\right)^2.$$ Indeed, after checking that this inequality is true for $a\rightarrow0^+$ and for $a\rightarrow+\infty$ and since continuous on a compact function gets a minimum

(let it happens in the point (a,b,c)), it's enough to check, what happens for $\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0,$ where $$f(a,b,c)=a\ln{a}+b\ln{b}+c\ln{c}-\frac{a+b+c}{1.5}\ln\frac{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}{3}.$$ Thus, $$\ln{a}+1-\frac{1}{1.5}\ln\frac{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}{3}-\frac{(a+b+c)\sqrt{a}}{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}=0,$$ $$\ln{b}+1-\frac{1}{1.5}\ln\frac{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}{3}-\frac{(a+b+c)\sqrt{b}}{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}=0$$ and $$\ln{c}+1-\frac{1}{1.5}\ln\frac{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}{3}-\frac{(a+b+c)\sqrt{c}}{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}=0.$$ We'll prove that it's possible for $a=b=c$ only.

Indeed, let it not be so and $a\leq b<c$.

If $a=b$ so let $a=b=xc$ and we need to prove that: $$(x^x)^2\geq\left(\frac{2\sqrt{x^3}+1}{3}\right)^{\frac{2x+1}{1.5}}$$ or $$x\geq\left(\frac{2\sqrt{x^3}+1}{3}\right)^{\frac{2x+1}{3x}}$$ $$x^2\geq\left(\frac{2x^3+1}{3}\right)^{\frac{2x^2+1}{3x^2}}$$ or $g(x)\geq0,$ where $$g(x)=2\ln{x}-\frac{2x^2+1}{3x^2}\ln\frac{2x^3+1}{3}.$$ which easy to show.

Indeed,$$g'(x)=\frac{2}{x}+\frac{2}{3x^3}\ln\frac{2x^3+1}{3}-\frac{2(2x^2+1)}{2x^3+1}=\frac{2}{3x^3}\left(\ln\frac{2x^3+1}{3}-\frac{3x^2(x-1)}{2x^3+1}\right),$$ $$\left(\ln\frac{2x^3+1}{3}-\frac{3x^2(x-1)}{2x^3+1}\right)_{x=1}=0$$ and $$\left(\ln\tfrac{2x^3+1}{3}-\tfrac{3x^2(x-1)}{2x^3+1}\right)'=\tfrac{3x(4x^4-2x^3-x+2)}{2x^3+1}=\tfrac{3x(2x^4+x+2(x-1)^2(x^2+x+1))}{2x^3+1}>0,$$ which says that $1$ is an unique root of $g'$ and since $$\lim_{x\rightarrow+\infty}\left(\ln\frac{2x^3+1}{3}-\frac{3x^2(x-1)}{2x^3+1}\right)=+\infty,$$ we obtain: $$g(x)\geq g(1)=0$$ for any $x>0$.

Now, let $a<b<c$.

Thus, we see that: $$\ln{b}-\ln{a}=\frac{(a+b+c)(\sqrt{b}-\sqrt{a})}{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}},$$ which gives $$\frac{\ln{b}-\ln{a}}{\sqrt{b}-\sqrt{a}}=\frac{\ln{c}-\ln{b}}{\sqrt{c}-\sqrt{b}}=\frac{\ln{c}-\ln{a}}{\sqrt{c}-\sqrt{a}}$$ or $$\frac{\ln\sqrt{b}-\ln\sqrt{a}}{\sqrt{b}-\sqrt{a}}=\frac{\ln\sqrt{c}-\ln\sqrt{b}}{\sqrt{c}-\sqrt{b}}=\frac{\ln\sqrt{c}-\ln\sqrt{a}}{\sqrt{c}-\sqrt{a}},$$ which is impossible by the Lagrange's mead value theorem.

Id est, it's enough to check $a=b=c$, which gives an equality.

Thus, it's enough to prove that: $$\left(\frac{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}{3}\right)^2\geq\sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$$ Now, let $a^3+b^3+c^3=3u$.

Thus, by Holder and AM-GM we obtain: $$\left(\frac{\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3}}{3}\right)^2=\frac{3u+2\sum\limits_{cyc}\sqrt{a^3b^3}}{9}=\frac{3u+2\sqrt{\frac{(1+1+1)\left(\sum\limits_{cyc}\sqrt{a^3b^3}\right)^2}{3}}}{9}\geq$$$$\geq\frac{3u+2\sqrt{\frac{(ab+ac+bc)^3}{3}}}{9}=\frac{u+2}{3}\geq\sqrt[3]{u}=\sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$$

Answer 2

It is not an answer, but rather some results which can be helpful for the further investigations.

1. $abc \leq 1$
   $\rhd$ $$\frac{ab +bc + ac}{3} \geq \sqrt[3]{a^2b^2c^2}\Rightarrow\sqrt[3]{a^2b^2c^2}\leq 1$$ $\lhd$

2. One can prove that $$\forall a_i > 0,\forall p_i>0: \sum\limits_{1}^{n}p_i = 1$$ the following inequality is fulfilled $$\sum\limits_{i =1}^{n}p_i a_i \geq \prod\limits_{i = 1}^{n}a^{p_i}_{i}.$$ Let us denote \begin{align} \frac{a}{a+b+c} &= \alpha\\ \frac{b}{a+b+c} &= \beta,\\ \frac{c}{a+b+c} &= \gamma, \end{align} then $\alpha+\beta+\gamma = 1$ and, according to the previous formula ($a_i = \frac1a, \frac1b, \frac1c$), $$(a^ab^bc^c)^{\frac{3}{a+b+c}} =(a^\alpha b^\beta c^\gamma)^3 \geq \Big(\frac{a+b+c}{3}\Big)^3 = \Big(\frac{a^3 + b^3 + c^3 + 9(a+b+c) - 3abc}{27}\Big)\geq \Big(\frac{a^3 + b^3 + c^3}{27} + \frac{a+b+c}{3} - \frac19\Big),$$ where the last inequality uses 1.