Question

[Joy]

x;y;z > 0 và x + y + z = 1

Tìm min T = 1/16x + 1/4y + 1/z

Answer 1

[Nguyễn Phương Uyên]

\(T=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\)  ; x + y + z = 1

\(\Rightarrow T=\frac{x+y+z}{16x}+\frac{x+y+z}{4y}+\frac{x+y+z}{z}\)

\(=\frac{1}{16}+\frac{y}{16x}+\frac{z}{16x}+\frac{x}{4y}+\frac{1}{4}+\frac{z}{4y}+\frac{x}{z}+\frac{y}{z}+1\)

\(=\left(\frac{1}{16}+\frac{1}{4}+1\right)+\left(\frac{y}{16x}+\frac{x}{4y}\right)+\left(\frac{z}{16x}+\frac{x}{z}\right)+\left(\frac{z}{4y}+\frac{y}{z}\right)\)                    (1)

\(x;y;z>0\Rightarrow\frac{y}{16x};\frac{x}{4y};\frac{z}{16x};\frac{x}{z};\frac{z}{4y};\frac{y}{z}>0\)

áp dụng bđt cô si : 

\(\frac{y}{16x}+\frac{x}{4y}\ge2\sqrt{\frac{y}{16x}\cdot\frac{x}{4y}}=\frac{1}{4}\)                             (2)

\(\frac{z}{16x}+\frac{x}{z}\ge2\sqrt{\frac{z}{16x}\cdot\frac{x}{z}}=\frac{1}{2}\)                                 (3)

\(\frac{x}{4y}+\frac{y}{z}\ge2\sqrt{\frac{z}{4y}\cdot\frac{y}{z}}=1\)                                        (4)

(1)(2)(3)(4) \(\Rightarrow T\ge\frac{1}{16}+\frac{1}{4}+1+\frac{1}{4}+\frac{1}{2}+1\)

\(\Rightarrow T\ge\frac{49}{16}\)

dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{y}{16x}=\frac{x}{4y}\\\frac{z}{16x}=\frac{x}{z}\\\frac{z}{4y}=\frac{y}{z}\end{cases}}\Leftrightarrow\hept{\begin{cases}4y^2=16x^2\\z^2=16x^2\\z^2=4y^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2x\\z=4x\\z=2y\end{cases}}\) có x+y+z = 1

=> x + 2x + 4x = 1

=> x = 1/7

xong tìm ra y = 2/7 và z = 4/7