Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}=\frac{1}{16x}+\frac{4}{16y}+\frac{16}{16z}\)
\(\ge\frac{\left(1+2+4\right)^2}{16\left(x+y+z\right)}=\frac{49}{16}\)
Dấu bằng xảy ra khi
\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
P = 1/16x + 1/4y + 1/z = (1/16x + 4/16y + 16/16z)
áp dụng BĐT Bunhiacopski ta có
(1/16x + 4/16y + 16/16z)*(16x + 16y + 16z) >= (1 + 2 + 4)^2 = 49
=> P.16 >= 49 hay P >= 49/16
dấu = xảy ra khi
1/(16x)^2 = 1/64y^2 = 1/16z^2 và x + y + z = 1
<> 1/16x = 1/8y = 1/4z và x + y + z = 1
<> 4x = 2y = z và x + y + z = 1
<> x = 1/7 và y = 2/7 và z = 4/7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{16}{z}\right)=\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2^2}{y}+\dfrac{4^2}{z}\right)\)
\(\Rightarrow M\ge\dfrac{1}{16}\dfrac{\left(1+2+4\right)^2}{x+y+z}=\dfrac{1}{16}.\dfrac{49}{1}=\dfrac{49}{16}\)
\(\Rightarrow M_{min}=\dfrac{49}{16}\) khi \(\left\{{}\begin{matrix}x+y+z=1\\\dfrac{1}{x}=\dfrac{2}{y}=\dfrac{4}{z}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{7}\\y=\dfrac{2}{7}\\z=\dfrac{4}{7}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{1}{16}\left(\dfrac{1}{x^2}+\dfrac{4}{y^2}+\dfrac{16}{z^2}\right)\ge\dfrac{1}{16}.\dfrac{\left(1+2+4\right)^2}{\left(x^2+y^2+z^2\right)}=\dfrac{49}{16}\)
\(\Rightarrow M_{min}=\dfrac{49}{16}\) khi \(\left\{{}\begin{matrix}x^2=\dfrac{1}{7}\\y^2=\dfrac{2}{7}\\z^2=\dfrac{4}{7}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}=\dfrac{1}{16x}+\dfrac{4}{16y}+\dfrac{16}{16z}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(M=\dfrac{1}{16x}+\dfrac{4}{16y}+\dfrac{16}{16z}=\dfrac{1^2}{16x}+\dfrac{2^2}{16y}+\dfrac{4^2}{16z}\)
\(\ge\dfrac{\left(1+2+4\right)^2}{16x+16y+16z}=\dfrac{7^2}{16\left(x+y+z\right)}=\dfrac{49}{16}\)
@Ace Legona tớ chưa học BĐT Cauchy-Schwarz ! Có cách giải khác không?
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{1}{16x^2}+\frac{1}{4y^2}+\frac{1}{z^2}\)
\(=\frac{1}{16x^2}+\frac{4}{16y^2}+\frac{16}{16z^2}\)
\(=\frac{1}{16}\left(\frac{1}{x^2}+\frac{4}{y^2}+\frac{16}{z^2}\right)\)
\(\ge\frac{1}{16}.\frac{\left(1+2+4\right)^2}{x^2+y^2+z^2}=\frac{49}{16}\)(Svac - xơ)
Vậy \(M_{min}=\frac{49}{16}\Leftrightarrow\frac{1}{x^2}=\frac{4}{y^2}=\frac{16}{z^2}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{\sqrt{21}}\\y=\frac{2}{\sqrt{21}}\\z=\frac{4}{\sqrt{21}}\end{cases}}\)
Cho sửa chỗ dấu "="
\("="\Leftrightarrow\frac{1}{x^2}=\frac{2}{y^2}=\frac{4}{z^2}=7\)
\(\Rightarrow\hept{\begin{cases}x=\sqrt{\frac{1}{7}}\\y=\sqrt{\frac{2}{7}}\\z=\frac{2}{\sqrt{7}}\end{cases}}\)hoặc \(\hept{\begin{cases}x=-\sqrt{\frac{1}{7}}\\y=-\sqrt{\frac{2}{7}}\\z=-\frac{2}{\sqrt{7}}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng BĐT Bunhiacopxky ta có:
\(\left(\frac{1}{16x^2}+\frac{1}{4y^2}+\frac{1}{z^2}\right)(x^2+y^2+z^2)\geq \left(\frac{1}{4}+\frac{1}{2}+1\right)^2\)
\(\Leftrightarrow M.1\geq \frac{49}{16}\Leftrightarrow M\geq \frac{49}{16}\)
Vậy \(M_{\min}=\frac{49}{16}\)
Dấu "=" xảy ra khi \((x,y,z)=(\sqrt{\frac{1}{7}}; \sqrt{\frac{2}{7}}; \sqrt{\frac{4}{7}})\)
Ta có : \(P=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}=\left(x+y+z\right)\left(\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}\right)\)( Vì \(x+y+z=1\) )
Áp dụng BĐT Bu - nhi - a - cốp - xki ta có :
\(\left(x+y+z\right)\left(\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}\right)\ge\left(\sqrt{x}.\dfrac{1}{4\sqrt{x}}+\sqrt{y}.\dfrac{1}{2\sqrt{y}}+\sqrt{z}.\dfrac{1}{\sqrt{z}}\right)^2=\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2=\dfrac{49}{16}\)
Dấu \("="\) xảy ra khi \(x=\dfrac{1}{7}\) ; \(y=\dfrac{2}{7}\) ; \(z=\dfrac{4}{7}\)
nếu không hiểu hỏi lại mình nhé!!!