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[laTEX]A = (1 + \frac{a+b+c}{a})(1 + \frac{a+b+c}{b})(1 + \frac{a+b+c}{c}) = ( 1+1 +\frac{b}{a}+\frac{c}{a}).(1+1 +\frac{a}{b}+\frac{c}{b}).(1+1 +\frac{a}{c}+\frac{b}{c}) [/laTEX]
áp dụng bất đẳng thức cosi với 4 số
[laTEX]1+1 +\frac{b}{a}+\frac{c}{a} \geq 4.\sqrt[4]{\frac{b.c}{a^2}} \\ 1+1 +\frac{a}{b}+\frac{c}{b} \geq 4.\sqrt[4]{\frac{c.a}{b^2}} \\ 1+1 +\frac{a}{c}+\frac{b}{c} \geq 4.\sqrt[4]{\frac{a.b}{c^2}}[/laTEX]
nhân 3 vế ta được
[laTEX]( 1+1 +\frac{b}{a}+\frac{c}{a}).(1+1 +\frac{a}{b}+\frac{c}{b}).(1+1 +\frac{a}{c}+\frac{b}{c}) \geq 4^3 = 64[/laTEX]
dấu = xảy ra khi
[laTEX]a= b = c = \frac{2003}{3}[/laTEX]