cho x,y,z >0 .và x+y+z=6
MAX P= $\frac{xy}{\sqrt{x ²+y ²+2z ²}}$ +$\frac{zy}{\sqrt{z ²+y ²+2x ²}}$ +$\frac{xz}{\sqrt{x ²+z²+2y ²}}$
cho x,y,z >0 .và x+y+z=6 MAX P= $\frac{xy}{\sqrt{x ²+y ²+2z ²}}$ +$\frac{zy}{\sqrt{z ²+y ²+2x ²}}$ +$\frac{xz}{\sqrt{x ²+z²+2y ²}}$
By Melody
`(xy)/(\sqrt(x^2+y^2+2z^2))≤ (xy)/(\sqrt(1/2((x+z)^2+(y+z)^2)))≤ (xy)/(\sqrt((x+z)(z+y)))≤\sqrt(xy)/\sqrt((x+z)) \sqrt(xy)/\sqrt((y+z))≤1/2 ((xy)/(x+z)+(xy)/(y+z))`
tương tự
`(xz)/(x^2+2y^2+z^2)≤1/2((xz)/(x+y)+(xz)/(y+z))`
`(zy)/(2x^2+y^2+z^2)≤ 1/2((yz)/(x+z)+(yz)/(y+x))`
`⇒P≤1/2 ((xy)/(x+z)+(xy)/(y+z)+(xz)/(x+y)+(xz)/(y+z)+(zy)/(x+z)+(bc)/(y+x))`
`⇒P≤1/2 ((y(x+z))/(x+z)+(x(y+z))/(y+z)+(z(x+y))/(x+y))`
`⇒P≤1/2 (x+y+z)`
`⇒P≤1/2 .6`
`⇒P≤3`
`”=”`xẩy ra khi :
`a=b=c=2`
vậy` maxP=3 `khi `a=b=c=2`
Đáp án:
Áp dụng bất đẳng thức cauchy ta có:
`x^2+z^2+z^2+y^2>=2\sqrt{(x^2+z^2)(z^2+y^2)}`
`<=>x^2+y^2+2z^2>=\sqrt{4(x^2+z^2)(y^2+z^2)}`
Áp dụng bất đẳng thức cauchy ta có:
`x^2+y^2+2z^2>=\sqrt{4(x^2+z^2)(y^2+z^2)}>=\sqrt{(x+z)^2(y+z)^2}=(x+z)(y+z)`
`=>(xy)/\sqrt{x^2+y^2+2z^2}<=(xy)/((x+z)(y+z))`
Áp dụng bất đẳng thức cauchy ta có:
`(xy)/\sqrt{x^2+y^2+2z^2}<=(xy)/((x+z)(y+z))<=1/2((xy)/(x+z)+(xy)/(y+z))`
CMTT:
`(yz)/\sqrt{z^2+y^2+2x^2}<=1/2((yz)/(x+y)+(xy)/(x+z))`
`(zx)/\sqrt{x^2+z^2+2y^2}<=1/2((xz)/(y+z)+(xz)/(x+y))`
Cộng từng vế các BĐT trên ta có:
`P<=1/2(x+y+z)=3`
Dấu “=” xảy ra khi `x=y=z=2`.