`x;y;z ≥0` `(xz)^2+(yz)^2+1 ≤3z` tìm`minP=1/((x+1)^2)+8/((y+3)^2)+(4z^2)/((1+2z)^2)`

0 bình luận về “`x;y;z ≥0` `(xz)^2+(yz)^2+1 ≤3z` tìm`minP=1/((x+1)^2)+8/((y+3)^2)+(4z^2)/((1+2z)^2)`”

1. `1/a^2+1/b^2 ≥4/(a^2+b^2)≥8/(2(a^2+b^2))≥8/(a+b)^2`

    `x^2z^2+y^2z^2+1≤3z`

   `⇔x^2+y^2+1/(z^2)≤3/z`

   `x^2+y^2+1/z^2+1+1+4≥2x+2y+4/z`

   mặt khác :

   `x^2+y^2+1/z^2+1+1+4≤3/z+6`

   `⇔3/z+6≥2x+2y+4/z`

   `⇔6≥1/z +2x+2y`

   `⇔P=1/(x+1)^2+8/(y+3)^2+(4z^2)/(1+2z)^2`

   `⇔P=1/(x+1)^2+8/(y+3)^2+(1)/(1/(2z)+1)^2≥8/(x+1/(2z)+2)^2+8/(y+3)^2≥(8^2)/(x+1/(2z)+y+5)^2≥(64)/((256)/(4))≥(64)/(64)=1`

   `”=”`xảy ra khi :
   `x=y=1`

   `z=1/2`

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