For $a,b,c>0$ satisfy $ab+bc+ca\ge11$. Prove that $$\sqrt[3]{a^2+3}+\frac{7}{5\sqrt[3]{14}}\sqrt[3]{b^2+3}+\frac{\sqrt[3]{9}}{5}\sqrt[3]{c^2+3}\ge \frac{23}{5\sqrt[3]{2}}$$
We have: $\sqrt[3]{a^2+3}=\sqrt[3]{4}*\sqrt[3]{\dfrac{\dfrac{a^2+1}{2}+1}{2}}\ge\sqrt[3]{4}*\dfrac{\sqrt[3]{\dfrac{a^2+1}{2}}+1}{2}$
$\sqrt[3]{b^2+3}=\sqrt[3]{7}*\sqrt[3]{\dfrac{5\dfrac{b^2+1}{5}+1+1}{7}}\ge\sqrt[3]{7}*\dfrac{5\sqrt[3]{\dfrac{b^2+1}{5}}+1+1}{7 }$
$\sqrt[3]{c^2+3}=\sqrt[3]{12}*\sqrt[3]{\dfrac{5\dfrac{c^2+1}{10}+1}{6}}\ge\sqrt[3]{12}*\dfrac{5\sqrt[3]{\dfrac{c^2+1}{10}}+1}{6}$
$\Rightarrow L.H.S\ge\dfrac{1}{\sqrt[3]{2}}\left(\sqrt[3]{\dfrac{a^2+1}{2}}+1\right)+\dfrac{1}{5\sqrt[3]{2}}\left(5\sqrt[3]{\dfrac{b^2+1}{5}}+2\right)+\dfrac{1}{5\sqrt[3]{2}}\left(5\sqrt[3]{\dfrac{c^2+1}{10}}+1\right)=\left(\sqrt[3]{\dfrac{a^2+1}{4}}+\sqrt[3]{\dfrac{b^2+1}{10}}+\sqrt[3]{\dfrac{c^2+1}{20}}\right)+\dfrac{8}{5\sqrt[3]{2}}$
By AM-GM $\sqrt[3]{\dfrac{a^2+1}{4}}+\sqrt[3]{\dfrac{b^2+1}{10}}+\sqrt[3]{\dfrac{c^2+1}{20}}\ge3\sqrt[9]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{800}}$
Need prove $3\sqrt[9]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{800}}+\dfrac{8}{5\sqrt[3]{2}}\ge\dfrac{23}{5\sqrt[3]{2}}$
$\Leftrightarrow (a^2+1)(b^2+1)(c^2+1)\ge100$
$\Leftrightarrow (a^2+1)((b+c)^2+(bc-1)^2)\ge100$
By C-S $\left(a^2+1\right)\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]\ge\left[a\left(b+c\right)+\left(bc-1\right)\right]^2=\left(ab+bc+ca-1\right)^2\ge10^2=100$
I need another way ( don't use AM-GM )
The first step we can make by Holder: $$\sqrt[3]{a^2+3}+\frac{7}{5\sqrt[3]{14}}\sqrt[3]{b^2+3}+\frac{\sqrt[3]{9}}{5}\sqrt[3]{c^2+3}=$$ $$=\tfrac{1}{\sqrt[3]4}\sqrt[3]{(1+1)^2(a^2+1+2)}+\tfrac{7}{5\sqrt[3]{14}}\cdot\tfrac{1}{\sqrt[3]{49}}\sqrt[3]{(5+2)^2(b^2+1+2)}+\tfrac{\sqrt[3]{9}}{5}\cdot\tfrac{1}{\sqrt[3]{144}}\sqrt[3]{(10+2)^2(c^2+1+2)}\geq$$ $$\geq\tfrac{1}{\sqrt[3]4}\left(\sqrt[3]{a^2+1}+\sqrt[3]2\right)+\tfrac{1}{5\sqrt[3]{2}}\left(\sqrt[3]{25(b^2+1)}+2\right)+\tfrac{1}{10\sqrt[3]2}\left(\sqrt[3]{100(c^2+1)}+2\right).$$ Id est, it's enough to prove that: $$\tfrac{1}{\sqrt[3]4}\left(\sqrt[3]{a^2+1}+\sqrt[3]2\right)+\tfrac{1}{5\sqrt[3]{2}}\left(\sqrt[3]{25(b^2+1)}+2\right)+\tfrac{1}{10\sqrt[3]2}\left(\sqrt[3]{100(c^2+1)}+2\right)\geq\frac{23}{5\sqrt[3]{2}}$$ or $$\sqrt[3]{\frac{a^2+1}{2}}+\sqrt[3]{\frac{b^2+1}{5}}+\sqrt[3]{\frac{c^2+1}{10}}\geq3,$$ which you made already.