Cho x,y,z > 0 và x+y+z = 1 . Tìm giá trị nhỏ nhất của biểu thức :
A = \(\dfrac{1}{x}\) + \(\dfrac{4}{y}\) + \(\dfrac{9}{z}\)
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Ta có \(\dfrac{1}{x+1}+\dfrac{1}{y+2}+\dfrac{1}{z+3}\ge\dfrac{9}{x+y+z+6}\), do đó:
\(\dfrac{9}{x+y+z+6}\le1\)
\(\Leftrightarrow x+y+z\ge3\)
Đặt \(x+y+z=t\left(t\ge3\right)\). Khi đó \(P=t+\dfrac{1}{t}\)
\(P=\dfrac{t}{9}+\dfrac{1}{t}+\dfrac{8}{9}t\)
\(\ge2\sqrt{\dfrac{t}{9}.\dfrac{1}{t}}+\dfrac{8}{9}.3\)
\(=\dfrac{2}{3}+\dfrac{24}{9}\)
\(=\dfrac{10}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}t=x+y+z=3\\x+1=y+2=z+3\end{matrix}\right.\)
\(\Leftrightarrow\left(x,y,z\right)=\left(2,1,0\right)\)
Vậy \(min_P=\dfrac{10}{3}\Leftrightarrow\left(x,y,z\right)=\left(2,1,0\right)\)
Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)
\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)
\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)
\(\dfrac{1}{x}+\dfrac{2}{y}\le1\Rightarrow\dfrac{2}{y}\le1-\dfrac{1}{x}\Rightarrow y\ge\dfrac{2x}{x-1}=2+\dfrac{2}{x-1}\)
\(x+\dfrac{2}{z}\le3\Rightarrow x< 3;\dfrac{2}{z}\le3-x\Rightarrow z\ge\dfrac{2}{3-x}\Rightarrow y+z\ge2+\dfrac{2}{x-1}+\dfrac{2}{3-x}\)
Lúc này ta sẽ áp dụng bất đẳng thức Bunhiacopski
Ta có:
\(6^2\le\left(y+z\right)^2=\left(\sqrt{2}\dfrac{y}{\sqrt{2}}Z\right)^2\le3\left(\dfrac{y^2}{2}+z^2\right)=\dfrac{3}{2}\left(y^2+2z^2\right)\)
\(\Rightarrow P\ge24\). Dấu đẳng thức xảy ra khi và chỉ khi \(y=4,z=2\)
Vậy giá trị nhỏ nhật của P là 24
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\Rightarrow (\frac{1}{x}+\frac{1}{y})+(\frac{1}{z}-\frac{1}{x+y+z})=0$
$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Leftrightarrow (x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)})=0$
$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y)(y+z)(x+z)=0$
$\Leftrightarrow x=-y$ hoặc $y=-z$ hoặc $z=-x$
Nếu $x=-y$ thì:
$P=\frac{3}{4}+[(-y)^8-y^8](y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}+0.(y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}$
Nếu $y=-z$ thì:
$P=\frac{3}{4}+(x^8-y^8)[(-z)^9+z^9](z^{10}-x^{10})=\frac{3}{4}+(x^8-y^8).0.(z^{10}-x^{10})=\frac{3}{4}$
Nếu $z=-x$ thì:
$P=\frac{3}{4}+(x^8-y^8)(y^9+z^9)[(-x)^{10}-x^{10}]=\frac{3}{4}+(x^8-y^8)(y^9+z^9).0=\frac{3}{4}$
Lời giải:
Sửa: $x^2\geq y^2+z^2$
Áp dụng BĐT Cauchy-Schwarz:
$P\geq \frac{y^2+z^2}{x^2}+\frac{7x^2}{2}.\frac{4}{y^2+z^2}+2007$
$=\frac{y^2+z^2}{x^2}+\frac{14x^2}{y^2+z^2}+2007$
$=\frac{y^2+z^2}{x^2}+\frac{x^2}{y^2+z^2}+\frac{13x^2}{y^2+z^2}+2007$
$\geq 2+\frac{13x^2}{y^2+z^2}+2007$ (áp dụng BĐT Cô-si)
$\geq 2+13+2007=2022$ (do $x^2\geq y^2+z^2$)
Vậy $P_{\min}=2022$
\(A=\dfrac{2x^2}{2x+2yz}+\dfrac{2y^2}{2y+2zx}+\dfrac{2z^2}{2z+2xy}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)
\(A\ge\dfrac{2x^2}{x^2+1+y^2+z^2}+\dfrac{2y^2}{y^2+1+z^2+x^2}+\dfrac{2z^2}{z^2+1+x^2+y^2}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)
\(A\ge\dfrac{2\left(x^2+y^2+z^2\right)}{x^2+y^2+z^2+1}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)
Đặt \(x^2+y^2+z^2=a>0\)
\(\Rightarrow A\ge\dfrac{2a}{a+1}+\dfrac{9}{8a}=\dfrac{2a}{a+1}+\dfrac{9}{8a}-\dfrac{15}{8}+\dfrac{15}{8}\)
\(\Rightarrow A\ge\dfrac{\left(a-3\right)^2}{8a\left(a+1\right)}+\dfrac{15}{8}\ge\dfrac{15}{8}\)
\(A_{min}=\dfrac{15}{8}\) khi \(a=3\) hay \(x=y=z=1\)
\(P=\dfrac{1}{2023}\dfrac{1}{z}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{2023.z}\dfrac{x+y}{xy}\)
Ap dung BDT cosi taco
\(P\ge\dfrac{1}{2023z}.\dfrac{x+y}{\dfrac{\left(x+y\right)^2}{4}}=\dfrac{4}{2023z}\dfrac{1}{x+y}\)
<->\(P\ge\dfrac{4}{2023}\dfrac{1}{z\left(1-z\right)}=\dfrac{4}{2023}\dfrac{1}{-z^2+z}=\dfrac{4}{2023}\dfrac{1}{-\left(z-\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)
\(< =>P\ge\dfrac{4}{2023}\dfrac{1}{\dfrac{1}{4}}=\dfrac{16}{2023}\)
\(P_{min}=\dfrac{16}{2023}\Leftrightarrow Z=\dfrac{1}{2},x=y=\dfrac{1}{4}\)
Áp dụng bđt Bunhiacopxki ta có :
\(A=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\right)\ge\left(\sqrt{x}.\dfrac{1}{\sqrt{x}}+\sqrt{y}.\dfrac{2}{\sqrt{y}}+\sqrt{z}.\dfrac{3}{\sqrt{z}}\right)^2\)
\(\left(1+2+3\right)^2=36\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel
\(A\ge\dfrac{\left(1+2+3\right)^2}{x+y+z}=36\)
Đẳng thức xảy ra khi \(x=\dfrac{1}{6};y=\dfrac{1}{3};z=\dfrac{1}{2}\)