Cho x,y,z > 0 thỏa mãn x2+y2+z2=1
Tìm giá trị nhỏ nhất để M = \(\dfrac{1}{16x^2}+\dfrac{1}{4y^2}+\dfrac{1}{z^2}\)
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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
\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)
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 $P_{\min}=2022$
\(M=\dfrac{1}{16}\left(\dfrac{1}{x^2}+\dfrac{4}{y^2}+\dfrac{16}{z^2}\right)\ge\dfrac{1}{16}.\dfrac{\left(1+2+4\right)^2}{\left(x^2+y^2+z^2\right)}=\dfrac{49}{16}\)
\(\Rightarrow M_{min}=\dfrac{49}{16}\) khi \(\left\{{}\begin{matrix}x^2=\dfrac{1}{7}\\y^2=\dfrac{2}{7}\\z^2=\dfrac{4}{7}\end{matrix}\right.\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
$A\geq \frac{9}{x+2+y+2+z+2}=\frac{9}{x+y+z+6}$
Áp dụng BĐT Bunhiacopxky:
$(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$
$\Rightarrow 9\geq (x+y+z)^2\Rightarrow x+y+z\leq 3$
$\Rightarrow A\geq \frac{9}{x+y+z+6}\geq \frac{9}{3+6}=1$
Vậy $A_{\min}=1$. Dấu "=" xảy ra khi $x=y=z=1$
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 \(min_P=\dfrac{10}{3}\Leftrightarrow\left(x,y,z\right)=\left(2,1,0\right)\)
Với mọi x;y;z ta luôn có:
\(\left(x+y-1\right)^2+\left(z-\dfrac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+2xy-2x-2y+1+z^2-z+\dfrac{1}{4}\ge0\)
\(\Leftrightarrow x^2+y^2+z^2+\dfrac{5}{4}+2xy-2x-2y-z\ge0\)
\(\Leftrightarrow2+2xy-2x-2y\ge z\)
\(\Leftrightarrow2\left(1-x\right)\left(1-y\right)\ge z\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{2}\)
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 \(x=\dfrac{1}{7}\) ; \(y=\dfrac{2}{7}\) ; \(z=\dfrac{4}{7}\)
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 \((x,y,z)=(\sqrt{\frac{1}{7}}; \sqrt{\frac{2}{7}}; \sqrt{\frac{4}{7}})\)