thac mac ve bai tap cua thay hinh

N

noinhobinhyen

ta biết rằng

$(x+y+z)^2 \leq 3(x^2+y^2+z^2)=3 => x+y+z \leq \sqrt{3}$

$xy+xz+yz \leq x^2+y^2+z^2=1$


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