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ngomaithuy93
[TEX]ab+bc+ca=abc \Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1[/TEX]cho a,b,c>0 . ab+bc+ca=abc
tìm max của P= (a+b-c-1)(b+c-a-1)(c+a-b-1) ?
[TEX]\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c} \Leftrightarrow a+b+c \leq 9[/TEX]
[TEX]P \leq (\frac{a+b+c-3}{3})^3=(\frac{a+b+c}{3}-1)^3[/TEX]
[TEX] f(t)=(\frac{t}{3}-1)^3--- t \in (-\infty;9][/TEX]
[TEX]\Rightarrow max f(t)=8 khi t=9[/TEX]
[TEX] \Rightarrow max P=8 \Leftrightarrow a=b=c=3[/TEX]
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