sunmoonlight
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How to prove that  > \dfrac{9}{a+b+c} )
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Inequality Proof (1 Viewer)
- Thread starter sunmoonlight
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Rising Renewal
by inspection 
liamkk112
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whats the restriction on a,b,c?How to prove that
WeiWeiMan
Well-Known Member
probably a,b,c > 0 and a≠b≠c since this looks like an application of AM/HM rearranged a bit
considerHow to prove that
(a+b+c)(1/a+1/b+1/c) > 9
(a+b)(1/a + 1/b) > 4
for LHS > MHS,
(a+b)(1/a+1/b) > 4
1/a + 1/b > 4/(a+b)
1/2(1/a + 1/b) > 2/(a+b)
similarly,
1/2(1/b+1/c) > 2/(b+c)
1/2(1/c+1/a) > 2/(c+a)
adding these, simplifying and factorising gives:
1/a + 1/b + 1/c > 2[1/(a+b) + 1/(b+c) + 1/(c+a)]
hence LHS > MHS
for MHS > RHS,
[(a+b)+(b+c)+(c+a)][1/(a+b)+1/(b+c)+1/(c+a)] > 9
2(a+b+c)[1/(a+b)+1/(b+c)+1/(c+a)] > 9
2[1/(a+b)+1/(b+c)+1/(c+a)] > 9/(a+b+c)
hence
(a+b)(1/a+1/b) > 4
1/a + 1/b > 4/(a+b)
1/2(1/a + 1/b) > 2/(a+b)
similarly,
1/2(1/b+1/c) > 2/(b+c)
1/2(1/c+1/a) > 2/(c+a)
adding these, simplifying and factorising gives:
1/a + 1/b + 1/c > 2[1/(a+b) + 1/(b+c) + 1/(c+a)]
hence LHS > MHS
for MHS > RHS,
[(a+b)+(b+c)+(c+a)][1/(a+b)+1/(b+c)+1/(c+a)] > 9
2(a+b+c)[1/(a+b)+1/(b+c)+1/(c+a)] > 9
2[1/(a+b)+1/(b+c)+1/(c+a)] > 9/(a+b+c)
hence
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