Find the minimum of $M=\frac{x}{y}+2\sqrt{1+\frac{y}{z}}+3\sqrt[3]{1+\frac{z}{x}}$

Question

Let $x,y,z>0$ such that $x\ge \max\{y,z\}$.Find the minimum of $$M=\dfrac{x}{y}+2\sqrt{1+\dfrac{y}{z}}+3\sqrt[3]{1+\dfrac{z}{x}}$$

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I found the $M_{min}=1+2\sqrt{2}+3\sqrt[3]{2}$ and $x=y=z$, then i don't have idea for the problem

Answer 1

$$M'(x)=\frac{1}{y}+\frac{-\frac{z}{x^2}}{\sqrt[3]{\left(1+\frac{z}{x}\right)^2}}=\frac{x^2\sqrt[3]{\left(1+\frac{z}{x}\right)^2}-yz}{y\sqrt[3]{\left(1+\frac{z}{x}\right)^2}}\geq0.$$ Thus, it's enough to check two cases:

1. $y\geq z$ and $x=y$.

Let $\frac{y}{z}=t$.

Hence, $t\geq1$ and we need to prove that $f(t)\geq0$, where $$f(t)=2\sqrt{1+t}+3\sqrt[3]{1+\frac{1}{t}}-2\sqrt2-3\sqrt[3]2.$$ Indeed, $$f'(t)=\frac{1}{\sqrt{1+t}}+\frac{-\frac{1}{t^2}}{\sqrt[3]{1+\frac{1}{t}}}=\frac{\sqrt[3]{t^4(t+1)^2}-\sqrt{1+t}}{\sqrt{1+t}\sqrt[3]{t^4(t+1)^2}}>0.$$ Thus, $f(t)\geq f(1)=0;$

2. $z\geq y$ and $x=z$.

Let $\frac{z}{y}=t$.

Hence, $t\geq1$ and we need to prove that $g(t)\geq0$, where $$g(t)=t+2\sqrt{1+\frac{1}{t}}-1-2\sqrt2.$$ Indeed $$g'(t)=1+\frac{-\frac{1}{t^2}}{\sqrt{1+\frac{1}{t}}}=\frac{\sqrt{t^3(t+1)}-1}{\sqrt{t^3(t+1)}}>0,$$ which says $g(t)\geq g(1)=0$.

Done!