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IF a+b+c=6 and a b c are positive real no. prove that 1/(6-a)+1/(6-b)+1/(6-c) is greater than equal to3/4

IF a+b+c=6 and a b c are positive real no. prove that 1/(6-a)+1/(6-b)+1/(6-c) is greater than equal to3/4

Grade:11

3 Answers

mycroft holmes
272 Points
7 years ago
Setting (6-a) = x, (6-b) = y, (6-c) = z, we have x+y+z = 12, x,y,z>0
 
From AM-GM we have (x+y+z) \left(\frac{1}{x} + \frac{1}{y} +\frac{1}{z} \right ) \ge 9
 
which means \left(\frac{1}{x} + \frac{1}{y} +\frac{1}{z} \right ) \ge \frac {9}{x+y+z} = \frac {3}{4}
 
The explanation for the inequality cited above can be found online or do let me know in case you need it
jagdish singh singh
173 Points
7 years ago
\hspace{-0.6 cm}$Using Cauchy-Schwarz Inequality::\\\\$\frac{1^2}{6-a}+\frac{1^2}{6-b}+\frac{1^2}{6-c}\geq \frac{(1+1+1)^2}{18-(a+b+c)}=\frac{9}{12}$\\\\\\And equality hold when $\frac{1}{6-a}=\frac{1}{6-b}=\frac{1}{6-c}=\frac{3}{12}$\\\\So equality hold when $a=b=c=2$
jagdish singh singh
173 Points
7 years ago
Very nice explanation hsbhatt sir

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