Minimize $S=\frac{a}{b} + 2\sqrt{\frac{b}{c}+1} + 3\sqrt[3]{\frac{c}{a}+1}$ under $a=\max\{a,b,c\}$ and $a,b,c>0$.

Question

Let $a, b, c > 0$ and $a \geqslant b, a \geqslant c$.Find the minimum value of $$S=\frac{a}{b} + 2\sqrt{\frac{b}{c}+1} + 3\sqrt[3]{\frac{c}{a}+1}.$$

From Diamonds in Mathmetical Inequalities, Tran Phuong

Answer 1

By AM-GM, $1+x\geq 2x^{\frac12}$ for all $x>0$, so $$S\geq \frac{a}{b}+2\sqrt{2}\left(\frac{b}{c}\right)^{\frac14}+3\sqrt[3]{2}\left(\frac{c}{a}\right)^{\frac{1}{6}}.$$ Using weighted AM-GM, we get $$S\geq (1+2\sqrt2+3\sqrt[3]2)\left(\left(\frac{a}{b}\right)\left(\frac{b}{c}\right)^{\frac{2\sqrt2}{4}}\left(\frac{c}{a}\right)^{\frac{3\sqrt[3]{2}}{6}}\right)^{\frac{1}{1+2\sqrt{2}+3\sqrt[3]{2}}}$$ or $$S\geq (1+2\sqrt2+3\sqrt[3]2)\left(\frac{a^{1-\frac{3\sqrt[3]2}{6}}}{b^{1-\frac{2\sqrt2}{4}}c^{\frac{2\sqrt2}{4}-\frac{3\sqrt[3]2}{6}}}\right)^{\frac{1}{1+2\sqrt{2}+3\sqrt[3]{2}}}\,.$$ Since $a\geq b$ and $a\geq c$, we obtain $$S\geq 1+2\sqrt2+3\sqrt[3]2.$$ It is easy to see that the equality holds iff $a=b=c$.