gtln

C

congchuaanhsang

Theo Cauchy 3 số:

$a+b+1$ \geq $3\sqrt[3]{ab}$

$b+c+1$ \geq $3\sqrt[3]{bc}$

$c+a+1$ \geq $3\sqrt[3]{ca}$

\Rightarrow $3P$ \leq $2(a+b+c)+3=9$

\Leftrightarrow $P$ \leq 3

$P_{max}=3$ \Leftrightarrow $a=b=c=1$
 
H

huynhbachkhoa23

Holder: $P^3 \le 9(ab+bc+ca) \le 3(a+b+c)^2 =27$

Suy ra $P \le 3$

Đẳng thức xảy ra khi $a=b=c=1$
 
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