Since $a+b+c =1$, we can write $$ (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1 +(ab+bc+ca) - abc = (ab+bc+ca) - abc = abc(a^-1 + b^-1 + c^-1 ) - abc = 3abc (fraca^-1 + b^-1 + c^-13) - abc.$$ What next?
Because $x+y ge 2sqrtxy$ for any $x>0,y>0$, and $1-a = (a+b+c) - a = b+c$ (similarly for $1-b$ và $1-c$) we have $$(1-a)(1-b)(1-c) = (b+c)(c+a)(a+b) ge 2sqrtbccdot 2sqrtcacdot 2sqrtab = 8abc.$$
You can rewrite $(1-a)(1-b)(1-c)$ as $abc(frac1a +frac1b + frac1c -1)$, meaning that your inequality is equivalent to lớn proving that $$frac 1a + frac 1b + frac 1c geq 9,$$
an inequality that is simpler to show.
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5xum"s answer here. Using $AMgeq HM$,$$$$$frac1a+frac1b+frac1cover3$$geq$$3overa+b+c$$$$$So, $frac1a+frac1b+frac1cgeq9$
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Show that the following inequality is true: $(frac1a + frac1bc) (frac1b + frac1ca)(frac1c + frac1ab) geq 1728$
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