1.

If `a+2b+3c=1` and ` a>0 , b>0 , c>0` and the greatest value of `a^3b^2c` is `1/k` then `(k-5180)` is

Answer» Here, `a+2b+3c = 1`
We can write it as,
`a/3+a/3+a/3+b+b+3c = 1`
Now, we know aritmatic mean of `n` numbers is always greater than geometric mean of those `n` numbers.
`:. (a/3+a/3+a/3+b+b+3c)/6 ge (a/3*a/3*a/3*b*b*3c)^(1/6)`
`=>(a+2b+3c)/6 ge ((a^3b^2c)/9)^(1/6)`
`=>1/6 ge ((a^3b^2c)/9)^(1/6)`
`=>(1/6)^6(9) ge (a^3b^2c)`
`=> a^3b^2c le 9/(216**216)`
`=> a^3b^2c le 1/5184`
Now, greatest value of ` a^3b^2c` is `1/k`,
`:. 1/k = 1/5184`
`=> k = 5184`
`:. k - 5180 = 5184-5180`
`=> k - 5180 = 4.`


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