Crazy Inequalities

Let's first compare $A$ and $B$ . Let's first assume $A > B$ . This means $a^ab^bc^c > a^ab^cc^b$ . This simplifies to $c^{c-b} > b^{c-b}$ , which is clearly true. Therefore $A > B$ .

Now let's compare $B$ with $C$ . Again, assume that $B > C$ . This means that $a^ab^cc^b > a^bb^cc^a$ . This simplifies to $c^{b-a} > a^{b-a}$ , which is true. Therefore $B > C$ .

Thus, the answer is $A > B > C$ , or $\boxed{C < B < A}$ .

Consider the ratio $\frac{A}{B}$ , which will exceed one if $A > B$ . Reducing the fraction, we find $\frac{A}{B} = \frac{b^bc^c}{b^cc^b} = b^{b - c}c^{c - b} = \left(\frac{c}{b}\right)^{c - b}.$ But $c > b$ , so $\frac{c}{b} > 1$ . Since the exponent $c - b$ is positive, we know $\frac{A}{B} = \left(\frac{c}{b}\right)^{c - b} > 1$ , and $A > B$ . Likewise, the ratio $\frac{B}{C}$ yields $\frac{B}{C} = \left(\frac{c}{a}\right)^{b - a} > 1,$ so $B > C$ .

Just assume any three value of a,b,c such as 0.1,0.2 and 0.3. Now use calculator. You get A>B>C or C<B<A.

A is the biggest because it contains the biggest possible power when using only a, b or c of c^c.

C 'wastes' base c, a big base on a, a small power, and wastes b as power on a as base. But B has a^a to stop this. So C is smaller than b.

The easiest method is to choose any real numbers on the interval. So, WLOG we can define a, b, and c to be rational.

Hence, let a=(1/8), b=(1/4), c=(1/2)

==> A = 1/(2^(1/8))^11

    B = 1/(2^(1/8))^13

    C = 1/(2^(1/8))^15

Taking ratios:

1) A/B = 2^(1/4)

2) A/C = 2^(1/2)

3) B/C = 2^(1/4)

We know that x^(1/2)>x^(1/4) for x > 1 .

Hence, 2^(1/2)>2^(1/4)

Also, x^(1/2) and x^(1/4) are > 1 for x >1

So, 2^(1/2)>2^(1/4)>1

==> A > C; A>B; B>C

Hence, A>B>C ==> C<B<A