Toán 9 Cho x,y,zx,y,z dương x+y+z=6x+y+z=6

[Hà Thanh kute]

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Cho x,y,z dương x+y+z=6
Tìm min P=x^3/√(2y^2+7yz+3z^2)+y^3/√(2z^2+7xz+3x^2)+z^3/√(2x^2+7xy+3y^2)

 

[shorlochomevn@gmail.com]

Cho x,y,z dương x+y+z=6
Tìm min P=x^3/√(2y^2+7yz+3z^2)+y^3/√(2z^2+7xz+3x^2)+z^3/√(2x^2+7xy+3y^2)

$$\frac{x^3}{\sqrt{2y^2+7yz+3z^2}}=\frac{x^3}{\sqrt{2y^2+yz+6yz+3z^2}}=\frac{x^3}{\sqrt{y.(2y+z)+3z.(2y+z)}}\\\\ =\frac{x^3}{\sqrt{(2y+z).(3z+y)}}\\\\ \frac{x^3}{\sqrt{(2y+z).(3z+y)}}+\frac{2}{3\sqrt{2}}.\sqrt{2y+z}+\frac{1}{\sqrt{6}}.\sqrt{3z+y}\geq 3\sqrt[3]{x^3.\frac{1}{\sqrt{3^3}}}\\\\ +, \sqrt{2y+z}.\sqrt{6}\leq \frac{2y+z+6}{2}=> \sqrt{2y+z}\leq \frac{2y+z+6}{2\sqrt{6}}\\\\ +,\sqrt{3z+y}.2\sqrt{2}\leq \frac{3z+y+8}{2}=> \sqrt{3z+y}\leq \frac{3z+y+8}{4\sqrt{2}}\\\\ => \frac{x^3}{\sqrt{(2y+z).(3z+y)}}\geq 3x\frac{1}{\sqrt{3}}-\frac{2}{3\sqrt{2}}.\sqrt{2y+z}-\frac{1}{\sqrt{6}}.\sqrt{3z+y}\\\\ \geq 3x\frac{1}{\sqrt{3}}-\frac{2}{3\sqrt{2}}.\frac{2y+z+6}{2\sqrt{6}}-\frac{1}{\sqrt{6}}.\frac{3z+y+8}{4\sqrt{2}}\\\\ =3x\frac{1}{\sqrt{3}}-\frac{2y+z+6}{6\sqrt{3}}-\frac{(3z+y+8)}{8\sqrt{3}}\\\\ => P\geq 3\frac{1}{\sqrt{3}}.(x+y+z)-\frac{1}{6\sqrt{3}}.[3.(x+y+z)+6.3]-\frac{1}{8\sqrt{3}}.[4.(x+y+z)+8.3]\\\\ => P\geq \frac{18}{\sqrt{3}}-\frac{6}{\sqrt{3}}-\frac{6}{\sqrt{3}}=2\sqrt{3}\\\\$$
dấu "=" <=> x=y=z=2

 

[NikolaTesla]

$$\frac{x^3}{\sqrt{2y^2+7yz+3z^2}}=\frac{x^3}{\sqrt{2y^2+yz+6yz+3z^2}}=\frac{x^3}{\sqrt{y.(2y+z)+3z.(2y+z)}}\\\\ =\frac{x^3}{\sqrt{(2y+z).(3z+y)}}\\\\ \frac{x^3}{\sqrt{(2y+z).(3z+y)}}+\frac{2}{3\sqrt{2}}.\sqrt{2y+z}+\frac{1}{\sqrt{6}}.\sqrt{3z+y}\geq 3\sqrt[3]{x^3.\frac{1}{\sqrt{3^3}}}\\\\ +, \sqrt{2y+z}.\sqrt{6}\leq \frac{2y+z+6}{2}=> \sqrt{2y+z}\leq \frac{2y+z+6}{2\sqrt{6}}\\\\ +,\sqrt{3z+y}.2\sqrt{2}\leq \frac{3z+y+8}{2}=> \sqrt{3z+y}\leq \frac{3z+y+8}{4\sqrt{2}}\\\\ => \frac{x^3}{\sqrt{(2y+z).(3z+y)}}\geq 3x\frac{1}{\sqrt{3}}-\frac{2}{3\sqrt{2}}.\sqrt{2y+z}-\frac{1}{\sqrt{6}}.\sqrt{3z+y}\\\\ \geq 3x\frac{1}{\sqrt{3}}-\frac{2}{3\sqrt{2}}.\frac{2y+z+6}{2\sqrt{6}}-\frac{1}{\sqrt{6}}.\frac{3z+y+8}{4\sqrt{2}}\\\\ =3x\frac{1}{\sqrt{3}}-\frac{2y+z+6}{6\sqrt{3}}-\frac{(3z+y+8)}{8\sqrt{3}}\\\\ => P\geq 3\frac{1}{\sqrt{3}}.(x+y+z)-\frac{1}{6\sqrt{3}}.[3.(x+y+z)+6.3]-\frac{1}{8\sqrt{3}}.[4.(x+y+z)+8.3]\\\\ => P\geq \frac{18}{\sqrt{3}}-\frac{6}{\sqrt{3}}-\frac{6}{\sqrt{3}}=2\sqrt{3}\\\\$$
dấu "=" <=> x=y=z=2

Mình ko hiểu 2 dòng này:

 

[shorlochomevn@gmail.com]

Mình ko hiểu 2 dòng này:
View attachment 134967

chứng minh tương tự 2 phân thức kia rồi cộng các vế....

 

[Hà Thanh kute]

$$\frac{x^3}{\sqrt{2y^2+7yz+3z^2}}=\frac{x^3}{\sqrt{2y^2+yz+6yz+3z^2}}=\frac{x^3}{\sqrt{y.(2y+z)+3z.(2y+z)}}\\\\ =\frac{x^3}{\sqrt{(2y+z).(3z+y)}}\\\\ \frac{x^3}{\sqrt{(2y+z).(3z+y)}}+\frac{2}{3\sqrt{2}}.\sqrt{2y+z}+\frac{1}{\sqrt{6}}.\sqrt{3z+y}\geq 3\sqrt[3]{x^3.\frac{1}{\sqrt{3^3}}}\\\\ +, \sqrt{2y+z}.\sqrt{6}\leq \frac{2y+z+6}{2}=> \sqrt{2y+z}\leq \frac{2y+z+6}{2\sqrt{6}}\\\\ +,\sqrt{3z+y}.2\sqrt{2}\leq \frac{3z+y+8}{2}=> \sqrt{3z+y}\leq \frac{3z+y+8}{4\sqrt{2}}\\\\ => \frac{x^3}{\sqrt{(2y+z).(3z+y)}}\geq 3x\frac{1}{\sqrt{3}}-\frac{2}{3\sqrt{2}}.\sqrt{2y+z}-\frac{1}{\sqrt{6}}.\sqrt{3z+y}\\\\ \geq 3x\frac{1}{\sqrt{3}}-\frac{2}{3\sqrt{2}}.\frac{2y+z+6}{2\sqrt{6}}-\frac{1}{\sqrt{6}}.\frac{3z+y+8}{4\sqrt{2}}\\\\ =3x\frac{1}{\sqrt{3}}-\frac{2y+z+6}{6\sqrt{3}}-\frac{(3z+y+8)}{8\sqrt{3}}\\\\ => P\geq 3\frac{1}{\sqrt{3}}.(x+y+z)-\frac{1}{6\sqrt{3}}.[3.(x+y+z)+6.3]-\frac{1}{8\sqrt{3}}.[4.(x+y+z)+8.3]\\\\ => P\geq \frac{18}{\sqrt{3}}-\frac{6}{\sqrt{3}}-\frac{6}{\sqrt{3}}=2\sqrt{3}\\\\$$
dấu "=" <=> x=y=z=2

Bạn giải giúp mk bài này với
Cm BĐT √a(4a+5b)+√b(4b+5a) <=3(a+b)

 

[shorlochomevn@gmail.com]

Bạn giải giúp mk bài này với
Cm BĐT √a(4a+5b)+√b(4b+5a) <=3(a+b)

$$A^2=(\sqrt{a.(4a+5b)}+\sqrt{b.(4b+5a)})^2\leq (a+b).(4a+5b+4b+5a)\\\\ => A^2\leq 9(a+b)^2\\\\ => A\leq 3.(a+b)$$
dấu "=" <=> .... <=> a=b

 

[Hà Thanh kute]

Giúp mk với ạ
Cho x+y=1
Tìm min N=1/(x^3+y^3)+1/xy

 

[shorlochomevn@gmail.com]

Giúp mk với ạ
Cho x+y=1
Tìm min N=1/(x^3+y^3)+1/xy

$$N=\frac{1}{x^3+y^3}+\frac{1}{xy}\\\\ =\frac{1}{(x+y)^3-3xy.(x+y)}+\frac{1}{xy}\\\\ =\frac{1}{1-3xy}+\frac{3}{3xy}\geq \frac{(1+\sqrt{3})^2}{1-3xy+3xy}=4+2\sqrt{3}$$
dấu "=" xảy ra <=> $$\frac{1}{1-3xy}=\frac{\sqrt{3}}{3xy}$$
và x+y=1
bạn tự giải dấu "=" nha... :>

 

[Hà Thanh kute]

Giúp mk với ạ
Giải hệ pt a) 2x^2-x(y-1)+y^2=3y
x^2+xy-3y^2=x-2y
b) x(x^2-y^2)=6y
2y(x^2+y^2)=5x

$$N=\frac{1}{x^3+y^3}+\frac{1}{xy}\\\\ =\frac{1}{(x+y)^3-3xy.(x+y)}+\frac{1}{xy}\\\\ =\frac{1}{1-3xy}+\frac{3}{3xy}\geq \frac{(1+\sqrt{3})^2}{1-3xy+3xy}=4+2\sqrt{3}$$
dấu "=" xảy ra <=> $$\frac{1}{1-3xy}=\frac{\sqrt{3}}{3xy}$$
và x+y=1
bạn tự giải dấu "=" nha... :>