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\(P+3=x+\left(y^2+1\right)+\left(z^3+1+1\right)\ge x+2y+3z\)
\(\Rightarrow P\ge x+2y+3z-3\)
\(6=\dfrac{1}{x}+\dfrac{4}{2y}+\dfrac{9}{3z}\ge\dfrac{\left(1+2+3\right)^2}{x+2y+3z}\)
\(\Rightarrow x+2y+3z\ge6\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng Bất đẳng thức: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (Tự chứng minh)
\(\Rightarrow C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
\(C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(Q=\dfrac{x^3}{y+z}+\dfrac{y^3}{x+z}+\dfrac{z^3}{x+y}\)
\(Q=\dfrac{x^4}{xy+xz}+\dfrac{y^4}{xy+zy}+\dfrac{z^4}{xz+yz}\)
\(Q\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+xz+xy+zy+xz+yz}=\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+xz\right)}\)(svac-xo)
Lại có:\(x^2+y^2+z^2\ge xy+yz+zx\)(tự cm)
\(\Rightarrow Q\ge\dfrac{x^2+y^2+z^2}{2}\)
Mặt khác:\(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\ge36\)(tự cm)
\(\Rightarrow x^2+y^2+z^2\ge12\)
\(\Rightarrow Q\ge\dfrac{12}{2}=6\)
Vậy MINQ=6<=>x=y=z=2
Ta có: \((\dfrac{x^3}{y+z}+\dfrac{y+z}{x})+\left(\dfrac{y^3}{x+z}+\dfrac{x+z}{y}\right)+\left(\dfrac{z^3}{x+y}+\dfrac{x+y}{z}\right)\ge2\sqrt{\dfrac{x^3\left(y+z\right)}{\left(y+z\right)x}}+2\sqrt{\dfrac{y^3\left(x+z\right)}{\left(x+z\right)y}}+2\sqrt{\dfrac{z^3\left(x+y\right)}{\left(x+y\right)z}}=2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=2\left(x+y+z\right)\ge2.6=12\)
(Bất đẳng thức cauchy)
mà \(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{y}{z} \)
\(=\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge2\sqrt{\dfrac{yx}{xy}}+2\sqrt{\dfrac{zx}{xz}}+2\sqrt{\dfrac{zy}{yz}}=2+2+2=6\) (Bất đẳng thức cauchy)
\(\Rightarrow P\ge12-6=6\)
Dấu "=" xảy ra \(\Leftrightarrow\)x = y = z = 2
Vậy GTNN của P = 6 \(\Leftrightarrow\)x = y = z = 2
\(A\ge\frac{9}{2x+y+2y+z+2z+x}=\frac{9}{3\left(x+y+z\right)}=\frac{9}{3.3}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Bài 1:
Vì $x+y+z=1$ nên:
\(Q=\frac{x}{x+\sqrt{x(x+y+z)+yz}}+\frac{y}{y+\sqrt{y(x+y+z)+xz}}+\frac{z}{z+\sqrt{z(x+y+z)+xy}}\)
\(Q=\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}}\)
Áp dụng BĐT Bunhiacopxky:
\(\sqrt{(x+y)(x+z)}=\sqrt{(x+y)(z+x)}\geq \sqrt{(\sqrt{xz}+\sqrt{xy})^2}=\sqrt{xz}+\sqrt{xy}\)
\(\Rightarrow \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:
\(Q\leq \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Vậy $Q$ max bằng $1$
Dấu bằng xảy ra khi $x=y=z=\frac{1}{3}$
Bài 2:
Vì $x+y+z=1$ nên:
\(\text{VT}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)
\(\text{VT}=\frac{(x+y+z)^2-x^2}{(x+y)(x+z)}+\frac{(x+y+z)^2-y^2}{(y+z)(y+x)}+\frac{(x+y+z)^2-z^2}{(z+x)(z+y)}\)
\(\text{VT}=\frac{(y+z)[(x+y)+(x+z)]}{(x+y)(x+z)}+\frac{(x+z)[(y+z)+(y+x)]}{(y+z)(y+x)}+\frac{(x+y)[(z+x)+(z+y)]}{(z+x)(z+y)}\)
Áp dụng BĐT AM-GM:
\(\text{VT}\geq \frac{2(y+z)\sqrt{(x+y)(x+z)}}{(x+y)(x+z)}+\frac{2(x+z)\sqrt{(y+z)(y+x)}}{(y+z)(y+x)}+\frac{2(x+y)\sqrt{(z+x)(z+y)}}{(z+x)(z+y)}\)
\(\Leftrightarrow \text{VT}\geq 2\underbrace{\left(\frac{y+z}{\sqrt{(x+y)(x+z)}}+\frac{x+z}{\sqrt{(y+z)(y+x)}}+\frac{x+y}{\sqrt{(z+x)(z+y)}}\right)}_{M}\)
Tiếp tục AM-GM cho 3 số trong ngoặc lớn, suy ra \(M\geq 3\)
Do đó: \(\text{VT}\geq 2.3=6\) (đpcm)
Dấu bằng xảy ra khi $3x=3y=3z=1$
Ta có:
\(=\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{y^2}{6}+\dfrac{y^2}{6}+\dfrac{y^2}{6}+\dfrac{z^3}{6}+\dfrac{z^3}{6}\)
\(\ge11.\sqrt[11]{\dfrac{x^6}{6^6}.\dfrac{y^6}{6^3}.\dfrac{z^6}{6^2}}=11.\sqrt[11]{\dfrac{\left(xyz\right)^6}{6^{11}}}=11.\sqrt[11]{\dfrac{1}{6^{11}}}=\dfrac{11}{6}\)
Vậy GTNN là \(A=\dfrac{11}{6}\)đạt được khi \(x=y=z=1\)
PS: Bài này nhé. Bài trước nhầm 1 chỗ. Mà kệ đừng xem bài trước làm gì nhé e.
Ta có:
\(=\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{x}{6}+\dfrac{y^2}{6}+\dfrac{y^2}{6}+\dfrac{y^2}{6}+\dfrac{z^3}{6}+\dfrac{z^3}{6}\)
\(\ge11.\sqrt[11]{\dfrac{x^6}{6^6}.\dfrac{y^6}{6^3}.\dfrac{z^6}{2^6}}=11.\sqrt[11]{\dfrac{\left(xyz\right)^6}{6^{11}}}=11.\dfrac{xyz}{6}=\dfrac{11}{6}\)
Vậy GTNN là \(A=\dfrac{11}{6}\)đạt được khi \(x=y=z=1\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)
\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{x+y+z}{xyz}\right)=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.1=3\) ( Do x+y+z=xyz )
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3-2=1\)
Vậy P = 1
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(6=\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{z}\)
\(\geq 6\sqrt[6]{\frac{1}{xy^2z^3}}\)
\(\Leftrightarrow \frac{1}{xy^2z^3}\leq 1\Leftrightarrow xy^2z^3\geq 1\)
Tiếp tục áp dụng BĐT AM-GM:
\(A=x+y^2+z^3\geq 3\sqrt[3]{xy^2z^3}\geq 3\sqrt[3]{1}=3\)
Vậy \(A_{\min}=3\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} \frac{1}{x}=\frac{1}{y}=\frac{1}{z}\\ x=y^2=z^3\end{matrix}\right.\Leftrightarrow x=y=z=1\)