The minimum value, may be without Calculus
{ a + 2 b + 3 c a 2 + b 2 + c 2 = = 1 2 9
Suppose a , b and c are real numbers fulfilling the above equations. Find the minimum value of ⌊ 1 0 0 0 c ⌋ .
Notation : ⌊ ⋅ ⌋ denotes the floor function .
The answer is -3000.
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2 solutions
Using the same trick as below, ( 7 a + 1 3 ) 2 + ( 7 b + 2 6 ) 2 + ( 7 c − 2 4 ) 2 = 3 0 2 4 − 8 8 2 c and hence c ≤ 7 2 4 , with equality when ( a , b , c ) = ( − 7 1 3 , − 7 2 6 , 7 2 4 ) .
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How did you come up with this form?? The terms 7 a + 1 3 ⋯ ???
Great solution +1!. Far better than mine. Just for the sake of variety, i mention my solution :-
I plugged in the variable a in terms of b and c from equation 1 to 2, getting ellipse with b and c as variables. Now finding lowest point of this ellipse did the job (this was tedious) .
Same solution +1
May I ask? Is cauchy applicable in 3 variables?
I used cauchy in 3 variables
i.e. (1+2+3)(a+b+c) <|= (a+2b+3c) => min (a+b+c) <|= 1/6 ?
Or how to solve this problem using AM-GM?
Note that ( a − 2 ) 2 + ( b − 4 ) 2 + ( c + 3 ) 2 = a 2 + b 2 + c 2 − 4 a − 8 b + 6 c + 2 9 = 5 8 − 4 ( a + 2 b + 3 c ) + 1 8 c = 5 4 + 1 8 c so that c = − 3 + 1 8 1 [ ( a − 2 ) 2 + ( b − 4 ) 2 + ( c + 3 ) 2 ] and hence the minimum value of c is − 3 , achieved with a = 2 , b = 4 . The answer is − 3 0 0 0 .
I get this one. Haha. The "cauchy inequality" isn't so familiar for me. Haha.
@Mark Hennings How did you come up with this form ???Why did you choose the specific numbers 2 , 4 , 3 ??
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It is basically the only way to express a sum of squares in terms of a 2 + b 2 + c 2 , a + 2 b + 3 c and c alone.
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Why can't we take this then ??
( a − 4 ) 2 + ( b − 8 ) 2 + ( c + 1 2 ) 2
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@Ankit Kumar Jain – Sorry, let me refine my previous comment. We want to be able to write ( a − α ) 2 + ( b − β ) 2 + ( c − γ ) 2 = k ( c − γ ) for constants α , β , γ and k > 0 .
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@Mark Hennings – ( a − 4 ) 2 + ( b − 8 ) 2 + ( c + 1 2 ) 2
= a 2 + b 2 + c 2 + 1 6 + 6 4 + 1 4 4 + 2 4 c − 8 a − 1 6 b
= 2 5 3 − 8 ( a + 2 b + 3 c ) + 4 8 c = 2 4 5 + 4 8 c , What is wrong with this?
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@Ankit Kumar Jain – Because the next step, of saying that c is minimized when all the squares vanish, is not longer true. When my expression involving c was c + 3 , both sides of the equation could be made zero at the same time. If we have the template I suggest, then c ≥ γ , and this is achieved when a = α , b = β and c = γ .
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@Mark Hennings – Yes you're right , I didn't notice that .
Thanks!!!
Using Cauchy-Schwartz Inequality, ( 1 + 4 ) ( a 2 + b 2 ) ≥ ( a + 2 b ) 2 . This implies that 5 ( 2 9 − c 2 ) = ( 1 − 3 c ) 2 . Expand, simplify and factorize it, we will obtain ( c + 3 ) ( 7 c − 2 4 ) ≤ 0 , which means that − 3 ≤ c ≤ 7 2 4 . Hence, c = − 3 is the minimum value and ⌊ 1 0 0 0 c ⌋ = − 3 0 0 0 . We should check that it is possible, take ( a , b , c ) = ( 2 , 4 , − 3 ) .
You may want to find the corresponding ( a , b , c ) if c is maximum.