If a>0, b>0, c>0, prove that?

a/b + b/c +c/a %26gt;= 3


Please more clearer.

If a%26gt;0, b%26gt;0, c%26gt;0, prove that?
ok more clearer.





let min be the minimum of a , b and c. ( the smallest of the three )





let max be the largest of a,b and c





then a/b %26gt;= min/b, because min is smaller (or equal ) than a.





next min/b %26gt;= min/max, because max is greater than b.


the bigger the number max, the smaller the ration min/max.





so netto you have that a/b %26gt;= min/max


this also holds fot the other 2 fractions.


thus :


a/b + b/c +c/a %26gt;= 3min/max





since min %26gt;= max min/max %26gt;= 1


thus


a/b + b/c +c/a %26gt;= 3min/max %26gt;= 3





Edit


this is simple , its just estimating downwards the ratios. i dont see a simpler now.





5/3 %26gt; 5/4 %26gt; 5/5 %26gt; 5/6 ... %26gt; 5/100


and 5/23 %26lt; 6/23 %26lt; 7/23 ... %26lt; 100/23


this is all that is used.
Reply:u cant prove that or atleast i cant
Reply:This is simple. Let a=b=c=1. Then you have 1/1+1/1+1/1=3.


Let a=b=c=0.01, 0.01/0.01+0.01/0.01+0.01/0.01=1+1+1=3.


Let a=b=1 and c=2. 1/1+1/2+2/1=1+0.5+2=3.5%26gt;3





The algebraic proof is too complex and also somewhat inconclusive This is due to the fact that there is no specified limit on whether or not a,b and c are less than 3 or not.
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Reply:a could be 1 b could be 2 and c could be 1