If a,b and c are three positive real numbers, which one of the following are true?
A
a2+b2+c2≥bc+ca+ab
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B
a3+b3+c3≥3abc
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C
(b+c)(c+a)(a+b)≥8abc
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D
bca+cab+abc≥a+b+c
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Solution
The correct options are A
a2+b2+c2≥bc+ca+ab
B
a3+b3+c3≥3abc
C
(b+c)(c+a)(a+b)≥8abc
D
bca+cab+abc≥a+b+c
We have a2+b22≥√a2b2=ab,b2+c22≥√b2c2=bcandc2+a22≥√c2a2=ca Adding these inequalities, we get a2+b2+c2≥bc+ca+abAlsoa3+b3+c33≥(a3b3c3)1/3⇒a3+b3+c3≥3abc Next,since(b+c)/2≥√bc,(c+a)/2≥√caand(a+b)/2≥√ab,weget(b+c2)(c+a2)(a+b2)≥√a2b2c2⇒(b+c)(c+a)(a+b)≥8abcLastly,wehave12(bca+cab)≥√bca.cab=c,12(cab+abc)≥√cab.abc=b,and12(abc+bca)≥√abc.bca=b Adding the above inequalities we get bca+cab+abc≥a+b+c