$0 \leq x,y,z \leq 2$, x+y+z=3. Tìm max $x^{2} + y^{2} + z^{2}$
$0 \leq x,y,z \leq 2$, x+y+z=3. Tìm max $x^{2} + y^{2} + z^{2}$
Started By AnAn333, 15-09-2023 - 15:32
Best Answer dinhvu, 16-09-2023 - 20:26
VT $=(x+y+z)^2-2(xy+yz+xz)=9-2(xy+yz+xz)$
Ta có$(x-2)(y-2)(z-2)\leq 0 \Rightarrow xyz+4x+4y+4z\leq 2xy+2yz+2xz+8\\ \Rightarrow 12 \leq 2xy+2yz+2xz+8\Rightarrow 2xy+2yz+2xz\geq 4\\ \Rightarrow x^2+y^2+z^2\leq 9-4=5$
Dấu "=" xảy ra chẳng hạn khi x=2,y=1,z=0
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#2
Posted 16-09-2023 - 20:26
dinhvu
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✓ Best Answer
VT $=(x+y+z)^2-2(xy+yz+xz)=9-2(xy+yz+xz)$
Ta có$(x-2)(y-2)(z-2)\leq 0 \Rightarrow xyz+4x+4y+4z\leq 2xy+2yz+2xz+8\\ \Rightarrow 12 \leq 2xy+2yz+2xz+8\Rightarrow 2xy+2yz+2xz\geq 4\\ \Rightarrow x^2+y^2+z^2\leq 9-4=5$
Dấu "=" xảy ra chẳng hạn khi x=2,y=1,z=0
Edited by dinhvu, 17-09-2023 - 13:34.
- William Nguyen and AnAn333 like this
#4
Posted 17-09-2023 - 13:33
dinhvu
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