Question

a,b and c are positive real numbers greater than or equal to 1 satisfying abc=100, and [a^(log a)]*[b^(log b)]*[c^(log c)]≥10000. What is the value of a+b+c?

Answer 1

well, i think you're just gonna have to do a lot of substitution

Answer 2

this is in brilliant :)

Answer 3

yep that makes sense. but how do I solve the part that requires log? :/

Answer 4

Lol this Reminds me of number 5 on the 2013 Part II AIME.

Answer 5

Pretty easy problem, think about it.

Answer 6

nah sorry I haven't covered logarithm in my syllabus, and though I read up on it, it didnt really work out :) thanks anyways!

Answer 7

Note the following:\[\bf abc=100 \implies \log(a)+\log(b)+\log(c)=2\]Whereever the base isn't given, assume it to be 10.

We can simplify the multiplication:\[\log(\bf a^{\log(a)}*b^{\log(b)}*c^{\log(c)}) \ge \log(10000) \]\[\implies \log(\bf a^{\log(a)})+\log(b^{\log(b)})+\log(c^{\log(c)}) \ge \log(10000) \]\[\implies [\log(\bf a)]^2+[\log(b)]^2+[\log(c)]^2 \ge 4\]Note that now we have enough information to use the formula:\[\bf (x+y+z)^2=x^2+y^2+z^2+2zx+2zy+2xy \]Where x = log(a), y = log(b), z = log(c). But note here that [log(a) + log(b) + log(c)]^2 = 4. From the previous inequality we know that the sum of squares of each logarithm is atleast 4. So if we substitude that in:\[\bf 4=4+2[\log(a)\log(b)+\log(a)\log(c)+\log(b)\log(c)] \]\[\bf \implies \log(a)\log(b)+\log(a)\log(c)+\log(b)\log(c)=0\]This statement here inplies that for the sum to be 0, we must have atleast one of the logarithms to be negative which means that atleast 1 of a,b, c has to be less than 1. But we are given the information that a, b, c are atleast equal to 1 which contradicts our result. This means that there is something wrong with either the question, or you haven't provided us with enough information, or you typed something up incorrectly.

@RaineJones

Answer 8

Another possibility is that I'm doing something wrong but I don't think that I am.

Answer 9

Thanks a lot! It's getting quite late though, so I'll probably complete this question tomorrow :) Thanks for the hints!

Answer 10

are you using \(\log~ a\) to mean \(\log_{10}(a)\)?

Answer 11

yup :)

Answer 12

Come on seriously? this one was easy too!

Answer 13

a,b and c are positive real numbers greater than or equal to 1 satisfying abc=100, and [a^(log a)]*[b^(log b)]*[c^(log c)]≥10000. What is the value of a+b+c?
$$abc=100\\\log abc=2\\\log a+\log b+\log c=2$$also$$a^{\log a}b^{\log b}c^{\log c}\ge10000\\\log(a^{\log a}b^{\log b}c^{\log c})\ge 4\\(\log a)^2+(\log b)^2+(\log c)^2\ge4$$

Answer 14

one of the numbers has to be 100

Answer 15

We know:$$(\log a+\log b+\log c)^2=4\\(\log a)^2+(\log b)^2+(\log c)^2\ge(\log a+\log b+\log c)^2$$

Answer 16

(in other words, \(\cdot^2\) is convex)

Answer 17

$$(\log a+\log b+\log c)^2-(\log a)^2+(\log b)^2+(\log c)^2\le0\\(\log a)(\log b)+(\log b)(\log c)+(\log a)(\log c)\le0\\AB+BC+AC\le 0$$Given that we know \(a,b,c\ge 1\) we find \(A,B,C\ge0\) and therefore without loss of generality we let trivial solution \(A=B=0\). But recall \(A+B+C=2\) so \(C=2\) and thus we find \(a+b+c=10^A+10^B+10^C=1+1+100=102\)

Answer 18

The idea is that we can let any two be \(0\) for the whole expression to reduce to \(0\).

Answer 19

Thanks a lot! :) @oldrin.bataku

Answer 20

I meant -(log x)^2 terms by the way in the first inequality of my second to last post