[Toán 9] Cho 3 số a,b,c thỏa mãn a+b+c =1

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p.i.e_66336

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noinhobinhyen

Có $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca \geq 3ab+3bc+3ca$

$\Rightarrow ab+bc+ca \leq \dfrac{1}{3}$

Dấu [=] xảy ra khi $x=y=z=\dfrac{1}{3}$
 
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