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[不等式] 证明托勒密不等式?

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hbghlyj Posted at 16 hr ago |Read mode
$\forall a,b,c,x,y,z\inR$
$$ \sqrt{(x-1)^2 + (y+1)^2 + (z+1)^2}\sqrt{(a-1)^2+(b+1)^2+(c+1)^2} $$
$$ + \sqrt{(x+1)^2 + (y-1)^2 + (z+1)^2}\sqrt{(a+1)^2+(b-1)^2+(c+1)^2} $$
$$ + \sqrt{(x+1)^2 + (y+1)^2 + (z-1)^2}\sqrt{(a+1)^2+(b+1)^2+(c-1)^2} $$
$$ \geq \sqrt{(x-1)^2 + (y-1)^2 + (z-1)^2}\sqrt{(a-1)^2+(b-1)^2+(c-1)^2} $$
即A=(1,−1,−1),B=(−1,1,−1),C=(−1,−1,1),D=(1,1,1),M=(x,y,z),N=(a,b,c),AM⋅AN+BM⋅BN+CM⋅CN≥DM⋅DN

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2025-4-21 19:11 GMT+8

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