Inequality Problem - Didn't use 2015 to avoid Complicacy!
a b c ( a 1 2 5 + b 1 2 5 + c 1 2 5 ) 1 6 ≤ K ( a 2 0 0 3 + b 2 0 0 3 + c 2 0 0 3 )
Find the value of the smallest constant K such that, for any positive real numbers a , b , c , the above inequality satisfies. Submit the value of ln ( K ) correct upto two decimal places as your answer.
Bonus: Can you find the smallest constant K in terms of ( p , q , n ) ∈ Z for the inequality below?
⎝ ⎛ i = 1 ∏ n x i ⎠ ⎞ ⎝ ⎛ i = 1 ∑ n x i p ⎠ ⎞ q ≤ K ⎝ ⎛ i = 1 ∑ n x i p q + n ⎠ ⎞
The answer is 16.48.
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with AM-GM inequality we get 3 ( a 2 0 0 3 + b 2 0 0 3 + c 2 0 0 3 ) ( ( a 3 + b 3 + c 3 ) ( a 2 0 0 0 + b 2 0 0 0 + c 2 0 0 0 ) ≥ 3 a b c ( a 2 0 0 0 + b 2 0 0 0 + c 2 0 0 0 ) (prove : expand) then set a 1 2 5 = x , b 1 2 5 = y , c 1 2 5 = z we want to find the best K for K ( x 1 6 + y 1 6 + z 1 6 ) ≥ ( x + y + z ) 1 6 . We use PM-AM 1 6 3 x 1 6 + y 1 6 + z 1 6 ≥ 3 x + y + z so K = 3 1 5