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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
\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)
\(x,y,z>0\)
Áp dụng BĐT Caushy cho 3 số ta có:
\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)
\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)
\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)
Áp dụng BĐT Caushy-Schwarz ta có:
\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)
\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)
\(P=0\Leftrightarrow x=y=z=1\)
Vậy \(P_{min}=0\)
\(M=\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{16}{z}\right)=\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2^2}{y}+\dfrac{4^2}{z}\right)\)
\(\Rightarrow M\ge\dfrac{1}{16}\dfrac{\left(1+2+4\right)^2}{x+y+z}=\dfrac{1}{16}.\dfrac{49}{1}=\dfrac{49}{16}\)
\(\Rightarrow M_{min}=\dfrac{49}{16}\) khi \(\left\{{}\begin{matrix}x+y+z=1\\\dfrac{1}{x}=\dfrac{2}{y}=\dfrac{4}{z}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{7}\\y=\dfrac{2}{7}\\z=\dfrac{4}{7}\end{matrix}\right.\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\right)(x+y+z)\geq \left(\sqrt{\frac{1}{16}}+\sqrt{\frac{1}{4}}+\sqrt{1}\right)^2\)
\(\Leftrightarrow P(x+y+z)\geq \frac{49}{16}\)
\(\Leftrightarrow P\geq \frac{49}{16}\) (do \(x+y+z=1\) )
Vậy \(P_{\min}=\frac{49}{16}\) tại \((x,y,z)=(\frac{1}{7}; \frac{2}{7}; \frac{4}{7})\)
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 \(x=\dfrac{1}{7}\) ; \(y=\dfrac{2}{7}\) ; \(z=\dfrac{4}{7}\)
\(M=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}=\dfrac{1}{16x}+\dfrac{4}{16y}+\dfrac{16}{16z}\)
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?
Đặt \(\dfrac{x}{z}=a;\dfrac{y}{z}=b\).
Theo gt ta có \(a+b\le1\).
BĐT cần chứng minh tương đương:
\(a^2+b^2+\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\ge \frac{21}{2}\).
Theo bđt AM - GM: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2;a^2+\dfrac{1}{16}a^2\ge\dfrac{1}{2};b^2+\dfrac{1}{16}b^2\ge\dfrac{1}{2};\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\ge\dfrac{15}{32}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\ge\dfrac{15}{32}.\left(\dfrac{4}{a+b}\right)^2\ge\dfrac{15}{2}\).
Cộng vế với vế của các bđt trên lại ta có đpcm.
\(M=\dfrac{1}{16}\left(\dfrac{1}{x^2}+\dfrac{4}{y^2}+\dfrac{16}{z^2}\right)\ge\dfrac{1}{16}.\dfrac{\left(1+2+4\right)^2}{\left(x^2+y^2+z^2\right)}=\dfrac{49}{16}\)
\(\Rightarrow M_{min}=\dfrac{49}{16}\) khi \(\left\{{}\begin{matrix}x^2=\dfrac{1}{7}\\y^2=\dfrac{2}{7}\\z^2=\dfrac{4}{7}\end{matrix}\right.\)
Áp dụng BĐT Shwarz:
\(M=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}=\dfrac{1}{16x}+\dfrac{4}{16y}+\dfrac{16}{16z}\)
\(\ge\dfrac{\left(1+2+4\right)^2}{16\left(x+y+z\right)}=\dfrac{49}{16}\)
Dấu " = " khi \(\dfrac{1}{16x}=\dfrac{2}{16y}=\dfrac{4}{16z}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{7}\\y=\dfrac{2}{7}\\z=\dfrac{4}{7}\end{matrix}\right.\)
Vậy...
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