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![](https://rs.olm.vn/images/avt/0.png?1311)
\(4=x+y+z\ge3\sqrt[3]{xyz}\Leftrightarrow\sqrt[3]{xyz}\le\dfrac{4}{3}\Leftrightarrow xyz\le\dfrac{64}{27}\)(BĐT cauchy)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{4}{3}\)
Lời giải:
Áp dụng BĐT AM-GM:
$xy\le \frac{(x+y)^2}{4}=\frac{(4-z)^2}{4}$
$\Rightarrow H\leq \frac{z(4-z)^2}{4}$
Tiếp tục áp dụng BĐT AM-GM:
$z(4-z)\leq \frac{(z+4-z)^2}{4}=4$
$4-z\leq 2$ do $z\geq 2$
$\Rightarrow \frac{z(4-z)^2}{4}\leq \frac{4.2}{4}=2$
Hay $H\leq 2$
Vậy $H_{\max}=2$ khi $(x,y,z)=(1,1,2)$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{\frac{1}{x+y+x}}=1\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\right)=1\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Leftrightarrow\left(x+y\right)\left[z\left(x+y+z\right)+xy\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
B=\(\left(x+y\right)\left(y+z\right)\left(z+x\right).M=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)
Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)
\(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
Cộng vế với vế ta có:
\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)
\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\sqrt{\left(x-3\right)^2+4^2}+\sqrt{\left(y-3\right)^2+4^2}+\sqrt{\left(z-3\right)^2+4^2}\)
\(P\ge\sqrt{\left(x-3+y-3+z-3\right)^2+\left(4+4+4\right)^2}=6\sqrt{5}\)
\(P_{min}=6\sqrt{5}\) khi \(x=y=z=1\)
Mặt khác với mọi \(x\in\left[0;3\right]\) ta có:
\(\sqrt{x^2-6x+25}\le\dfrac{15-x}{3}\)
Thật vậy, BĐT tương đương: \(9\left(x^2-6x+25\right)\le\left(15-x\right)^2\)
\(\Leftrightarrow8x\left(3-x\right)\ge0\) luôn đúng
Tương tự: ...
\(\Rightarrow P\le\dfrac{45-\left(x+y+z\right)}{3}=14\)
\(P_{max}=14\) khi \(\left(x;y;z\right)=\left(0;0;3\right)\) và hoán vị
![](https://rs.olm.vn/images/avt/0.png?1311)
Tham khảo:
Cho 3 số thức x,y,z thỏa mãn \(x\ge1;y\ge4;z\ge9\) tìm giá trị lớn nhất của biết thức Q=\(\dfrac{yz\sqrt{x-1}+zx\sqrt... - Hoc24
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+z^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge2+2+2=6\)(BDT cô-si)
Dấu '=' xảy ra khi x=y=z=1 rồi thay vào tính dc P=3
\(x^2+y^2+z^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=6\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)+\left(z^2+\frac{1}{z^2}-2\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2+\left(z-\frac{1}{z}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\\z-\frac{1}{z}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=1\\z^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm1\\y=\pm1\\z=\pm1\end{cases}}\)
=> \(P=x^{28}+y^{10}+z^{2017}=1+1+z^{2017}=2+z^{2017}\)
Với \(z=-1\Rightarrow P=1+1-1=1\)
Với \(z=1\Rightarrow P=1+1+1=3\)
Ap dung BDT cosi,ta co:
3x^2017+2014=x^2017+x^2017+x^2017+1+1+............+1>=2017.x^3
CMTT
suy ra 2017(x^3+y^3+z^3)<=3(x^2017+y^2017+z^2017+2014)=6051
Suy ra max= 6051,dau bang xay ra khi va chi khi x=y=z=1