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If a,b,c are distinct positive real numbers and a 2 b 3 c 5 =2 10 ,then 2a+3b+5c is always greater than a.10 b.20 c.1 d.5

If a,b,c are distinct positive real numbers and a2b3c5 =210,then 2a+3b+5c is always greater than


a.10


b.20


c.1


d.5

Grade:12th Pass

4 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Menka,

 

Us AM-GM Ineguality on 2As,3Bs, and 5Cs.

 

So (2A+3B+5C)/10 > [A2B3C5]1/10

Hence, 2A+3B+5C > 20.

 

Hope it helps.

 

Regards,

Ashwin (IIT Madras).

Swapnil Saxena
102 Points
9 years ago

Let a,b,c Be three different real no. such that  a2b3c5 =210

Putting log2 on both sides log2 (a2b3c5) =log2 210

= log2 (a2)+  log2(b3)+ log2(c5) =log2 (210)

= 2 log2 (a) + 3 log2(b)+ 5 log2(c) =10

= As per the general rule x > log2(x) ( for all positive real no. this rule is followed), then  2 (a) + 3 (b)+ (c) > 10 as a> log2(a) , b> log2(b) , c> log2(c)

So (a.) must be the correct answer.

Swapnil Saxena
102 Points
9 years ago

Sorry , option (b) is correct according to the AM-GM inequality for positive no.

2A+3B+5C = (A+A) +(B+B+B)+(C+C+C+C+C)

According to the AM-GM inequalty,

(x1+x2+x3+x4+...xn)/n > (x1x2x3...xn)1/n

So (A+A) +(B+B+B)+(C+C+C+C+C) consist of 10 terms thus

((A+A) +(B+B+B)+(C+C+C+C+C))/10 > ((A*A) *(B*B*B)+(C*C*C*C*C))1/10

So (2A+3B+5C)/10 > ((A2) *(B3)+(C5))1/10

thus (2A+3B+5C) /10 > (210)1/10

(2A+3B+5C) > (20) -----(Ans)

ankitesh gupta
63 Points
8 years ago

 

 

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