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noinhobinhyen

ta biết rằng

(x+y+z)2≤3(x2+y2+z2)=3=>x+y+z≤3(x+y+z)^2 \leq 3(x^2+y^2+z^2)=3 => x+y+z \leq \sqrt{3}

xy+xz+yz≤x2+y2+z2=1xy+xz+yz \leq x^2+y^2+z^2=1


=>MAXP=3+1<=>x=y=z=13=> MAX_P = \sqrt{3}+1 <=> x=y=z=\dfrac{1}{\sqrt{3}}
 
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