minimum value?

Question

a,b and c are three positive real numbers . The minimum value that the expression (a+b)/ 2 x (b+c)/2 x (c+a)/2 can take when the product of the three numbers is 3/2 is

(1)3/2
(2)3
(3)8
(4)1

How to solve it ?

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Hi, you have made a mistake ...there is a multiplication .

Answer 1

The expression can expand to

[a^2 *(b+c) + b^2*(a+c) + c^2*(a+b) + 2abc ] /8

we know abc = 3/2, and all 3 are positive

So we have
[a^2 *(b+c) + b^2*(a+c) + c^2*(a+b) + 3 ] /8
This is a positive number greater than 3/8

Let's look at
a^2 *(b+c) + b^2*(a+c) + c^2*(a+b)
= ab*(a+b) + ac(a+c) + bc(b+c)
= (3/2)* [(a+b)/c + (a+c)/b + (b+c)/a ]

If either of a, b, or c is very small, then this term will become very large. We want the minimum.
Let's look at the case where a = b = c = 1.5^1/3
we get (3/2) * (6) = 9

So the full expression becomes
[a^2 *(b+c) + b^2*(a+c) + c^2*(a+b) + 3 ] /8
= [9 + 3 ] /8
= 1.5

I'd go for 3/2 = 1.5.
Those functions are usually optimized when a, b and c are close to each other.

Answer 2

ANSWER 3/2

We are looking at the avergae of all the numbers!
a+b / 2 ==> avg of a and b
b+c / 2 ===> avg of b and c
a+c / 2 ===> avg of a and c

So, if we minimize each number in the product, then we will minimize the sum of their averages.

We are just minimizing a*b*c
if they are close together, then a=b=c ==>a^3 = 3/2
a is about=to cuberoot of 3/2
average of these three numbers will be about cuberoot3/2 (so close together, they will equal each other)

cuberoot3/2 * cuberoot3/2 * cuberoot3/2 = 3/2

ANSWER 3/2


Previously I had said:
1*3/2*1
5/4 + 5/4 + 1 = 14/4 (CAN ONLY BE 3/2 or 1)