{ a + b + c + d = 1 2 a b c d = 2 7 + a b + a c + a d + b c + b d + c d
If the above equations hold true simultaneously for some positive real numbers a , b , c and d , then find the value of ( 2 . d 1 4 a 4 . b 5 . c 6 )
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exactly ! nice solution @DEEPANSHU GUPTA .. Congrats for being Moderator. You are welcome. :)
This is a question in Arihant Integral Calculus isn't it?
Exactly what I did..Upvoted! Minor typo : It should be RHS ≥ 2 7 + 6 a b c d .
when I was typing my answer 1.5 it not responded
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It is Good Application of AM-GM.
Since We have 4 variables and 2 equations So it can't solved directly. And all 4 variables are positive real variables. So first Thing that must strike in our Mind is that is AM-GM inequality can be used or not ?
( I mean Is there is boundedness of cool variables or not ? )
Let's Take a try:
∵ a + b + c + d = 1 2 ∴ 4 a + b + c + d ≥ ( a b c d ) 4 1 ∴ ( a b c d ) m a x = 8 1 .
Now look to the second Information:
∵ a . b . c . d = 2 7 + a . b + a . c + a . d + b . c + b . d + c . d L H S ≤ 8 1 ⟶ ( 1 ) .
Similarly By using AM-GM inequality on variables Of RHS
R H S ≥ 2 7 + 6 ( 4 3 a b c d ) R H S ≥ 8 1 ⟶ ( 2 ) .
But LHS=RHS=81 only possible solution.
So AM=GM and therefore numbers are equal .
a = b = c = d = 3 .
Now Put the Value and get the Answer.
It is interesting that we have given 4 variables and 2 equation's but still they can be solved By using simple AM-GM inequality.