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Question

cho x, y, z >0 thỏa mãn x+y+z=1

chứng minh rằng :$\dfrac{3}{xy+yz+xz}+\dfrac{2}{x^{2^{ }}+y^{2^{ }}+z^{2^{ }}}$≥14

Feed Là Quyền Công Dân · Accepted Answer

$VT=\dfrac{3}{xy+yz+xz}+\dfrac{2}{x^2+y^2+z^2}$

$=\dfrac{8}{4\left(xy+yz+xz\right)}+\dfrac{4}{4\left(xy+yz+xz\right)}+\dfrac{4}{2\left(x^2+y^2+z^2\right)}$

$\ge\dfrac{8}{4\cdot\dfrac{\left(x+y+z\right)^2}{3}}+\dfrac{\left(2+2\right)^2}{2\left(x+y+z\right)^2}$

$=\dfrac{8}{4\cdot\dfrac{1^2}{3}}+\dfrac{\left(2+2\right)^2}{2\cdot1^2}=14$

$"="\Leftrightarrow x=y=z=\dfrac{1}{3}$Câu trả lời: $VT=\dfrac{3}{xy+yz+xz}+\dfrac{2}{x^2+y^2+z^2}$

$=\dfrac{8}{4\left(xy+yz+xz\right)}+\dfrac{4}{4\left(xy+yz+xz\right)}+\dfrac{4}{2\left(x^2+y^2+z^2\right)}$

$\ge\dfrac{8}{4\cdot\dfrac{\left(x+y+z\right)^2}{3}}+\dfrac{\left(2+2\right)^2}{2\left(x+y+z\right)^2}$

$=\dfrac{8}{4\cdot\dfrac{1^2}{3}}+\dfrac{\left(2+2\right)^2}{2\cdot1^2}=14$

$"="\Leftrightarrow x=y=z=\dfrac{1}{3}$

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