Cho x,y,z dương thỏa mãn $x+y+z=1.$CMR:$44(xy+yz+xz)/ \leq (3x+4y+5z)^2$
:$44(xy+yz+xz) \leq (3x+4y+5z)^2$
Started By sharker, 15-04-2017 - 05:54
#2
Posted 15-04-2017 - 12:57
$(3x+4y+5z)^2-44(xy+yz+zx)=9x^2-20xy+16y^2-4yz+25z^2-14zx$
$=(\frac{20}{3}x^2-20xy+15y^2)+(y^2-4yz+4z^2)+(21z^2-14zx+\frac{7}{3}x^2)$
$=\frac{5}{3}(2x-3y)^2+(y-2z)^2+\frac{7}{3}(3z-1)^2\geqslant 0$
$\Rightarrow (3x+4y+5z)^2\geqslant 44(xy+yz+zx)$
Dấu "=" xảy ra khi $x=\frac{1}{2},y=\frac{1}{3},z=\frac{1}{6}$
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#3
Posted 16-04-2017 - 17:10
Cho x,y,z dương thỏa mãn $x+y+z=1.$CMR:$44(xy+yz+xz)/ \leq (3x+4y+5z)^2$
Ta có
\[(3x+4y+5z)^2 - 44(xy+yz+xz) = \frac19(9x-10y-7z)^2+\frac{44}{9}(y-2z)^2 \geqslant 0.\]
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Nguyen Van Huyen
Ho Chi Minh City University Of Transport
Ho Chi Minh City University Of Transport
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