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RE: Can you solve?

in #math6 years ago

In order to simplify the expression, let’s set a + b = p and a+b+c = q

Then we have:

Combining the last two fractions:

If I rearrange the above slightly, I get:

If I continue the substitutions:

Now this contains all positive terms. The "greater or equal than" tells me I should attempt to find a minimum value, and to do that I need to introduce a negative term somehow. I do that by:

To make the numerator smaller, you want to maximize ab, which requires that a = b. Similarly, this also requires that c = b.

Therefore, (1/2)+(1/2)+(1/2) = 3/2 is the minimum value of the expression.

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Good try so far!

a, b, & c are all independent variables. Note that a & b are (positive and) arbitrarily large, which means they are not necessarily maximized when they are equal. Indeed, for any choice of a, there are infinitely many b that make ab > a*a.

But here's a hint: You can make certain choices and not alter the independence of the 3 variables.

Thanks for at least trying! You made my week. :)

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