Let \$a\$, \$b\$, và \$c\$ be positive real numbers such that \$a+b+c = 1\$. Then how khổng lồ establish the following inequality? \$\$ (1-a)(1-b)(1-c) geq 8abc.\$\$

My effort:

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.

Bạn đang xem: Prove that ` 5xum"s answer here. Using \$AMgeq HM\$,\$\$\$\$\$frac1a+frac1b+frac1cover3\$\$geq\$\$3overa+b+c\$\$\$\$\$So, \$frac1a+frac1b+frac1cgeq9\$ But avoid

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