Question

[Gió]

Cho \(\left\{{}\begin{matrix}x,y,z>0\\x\ge max\left\{y,z\right\}\end{matrix}\right.\). Tìm Min của:

\(M=\dfrac{x}{y}+2\sqrt{1+\dfrac{y}{z}}+3\sqrt[3]{1+\dfrac{z}{x}}\)

P/s: Đề trc bị sai nhé!

Answer 1

[Serena chuchoe]

Áp dụng bđt AM-GM có:

\(1+\dfrac{y}{z}\ge2\sqrt{\dfrac{y}{z}};1+\dfrac{z}{x}\ge2\sqrt{\dfrac{z}{x}}\)

Dễ dàng suy ra: \(M\ge\dfrac{x}{y}+2\sqrt{2}\cdot\sqrt[4]{\dfrac{y}{z}}+3\sqrt[3]{2}\cdot\sqrt[6]{\dfrac{z}{x}}=\dfrac{1}{\sqrt{2}}\left(\dfrac{x}{y}+4\sqrt[4]{\dfrac{y}{z}}+6\sqrt[6]{\dfrac{z}{x}}\right)+\left(1-\dfrac{1}{\sqrt{2}}\right)\cdot\dfrac{x}{y}+\left(3\sqrt[3]{2}-3\sqrt{2}\right)\cdot\sqrt[6]{\dfrac{z}{x}}\)

Theo AM-GM có: \(\dfrac{1}{\sqrt{2}}\left(\dfrac{x}{y}+4\sqrt[4]{\dfrac{y}{z}}+6\sqrt[6]{\dfrac{z}{x}}\right)\ge\dfrac{1}{2}\cdot11\sqrt[11]{\dfrac{x}{y}\cdot\dfrac{y}{z}\cdot\dfrac{z}{x}}=\dfrac{11}{\sqrt{2}}\) (1)

Theo đề: \(x\ge max\left\{y,z\right\}\) ta có: \(\left\{{}\begin{matrix}\dfrac{x}{y}\ge1\\\dfrac{z}{x}\le1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left(1-\dfrac{1}{\sqrt{2}}\right)\cdot\dfrac{x}{y}\ge1-\dfrac{1}{\sqrt{2}}\left(2\right)\\\left(3\sqrt[3]{2}-3\sqrt{2}\right)\cdot\sqrt[6]{\dfrac{z}{x}}\ge3\sqrt[3]{2}-3\sqrt{2}\left(3\right)\end{matrix}\right.\)

Cộng theo vế bđt (1), (2) ,(3) có:\(A\ge\dfrac{11}{\sqrt{2}}+1-\dfrac{1}{\sqrt{2}}+3\sqrt[3]{2}-3\sqrt{2}=1+2\sqrt{2}+3\sqrt[3]{2}\)

Xảy ra khi \(x=y=z\)

Answer 2

[Lightning Farron]

Lâu lâu k đi khủng bố tinh thần :3

Ta đi cm \(1+2\sqrt{2}+3\sqrt[3]{2}\) là Min nhé

\(M'(x)=\dfrac{1}{y}+\dfrac{-\dfrac{z}{x^2}}{\sqrt[3]{\left(1+\dfrac{z}{x}\right)^2}}=\dfrac{x^2\sqrt[3]{\left(1+\dfrac{z}{x}\right)^2}-yz}{y\sqrt[3]{\left(1+\dfrac{z}{x}\right)^2}}\ge0\)

Vì vậy ta cần xét 2 trường hợp

*)\(y\ge z;x=y\). Đặt \(\dfrac{y}{z}=t\). Khi đó \(t\ge 1\) và cần cm \(f(t)\ge 0\)

\(f(t)=2\sqrt{1+t}+3\sqrt[3]{1+\dfrac{1}{t}}-2\sqrt{2}-3\sqrt[3]{2}\)

Thật vậy \(f'(t)=\dfrac{1}{\sqrt{1+t}}+\dfrac{-\dfrac{1}{t^2}}{\sqrt[3]{1+\dfrac{1}{t}}}=\dfrac{\sqrt[3]{t^4(t+1)^2}-\sqrt{1+t}}{\sqrt{1+t}\sqrt[3]{t^4(t+1)^2}}>0\)

\(\Rightarrow f(t)\ge f(1)=0\)

*)\(z\ge y ;x=z\). Khi đó \(t\ge 1\) và ta cm \(g(t)\ge 0\)

\(g(t)=t+2\sqrt{1+\dfrac{1}{t}}-1-2\sqrt{2}\)

Và \(g'(t)=1+\dfrac{-\dfrac{1}{t^2}}{\sqrt{1+\dfrac{1}{t}}}=\dfrac{\sqrt{t^3(t+1)}-1}{\sqrt{t^3(t+1)}}>0\)

Tức là \(g(t)\geq g(1)=0\)