[Toán 9] Cực trị

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huytrandinh

BỔ ĐỀ
[TEX](x+y+z)^{2}\geq 3(xy+yz+xz)[/TEX]
[TEX]<=>(x-y)^{2}+(y-z)^{2}+(x-z)^{2}\geq 0(dpcm)[/TEX]
[TEX]=>xy+yz+xz\leq \frac{1}{3}(x+y+z)^{2}=3[/TEX]
[TEX]=>max=3<=>x=y=z=1[/TEX]
 
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