Math  Algebra
posted by Momo .
Solve the inequality.
n1/3n < n4/3n4
(it's actually less than or equal to)
My first instinct is to multiply 3n, but that'll just give me squares and makes it harder to solve. I also thought about getting a common denominator, but I'm not sure.
Thank you!

I assume you mean
(n1)/3n <or= (n4)/(3n4)
When you multiply or divide both sides of inequalities, you have to pay attention to the sign of what you are multiplyiong or dividing by. We need to find out what n is first, when the two sides are equal.
Consider the e"=" case first. The equation leads to
3n(n4) = (n1)(3n4)
3n^2 12n = 3n^2 7n +4
The n^2 terms cancel out, and
5n = 4
n = 4/5
Now go back to the original inequality. If you multiply both sides by (3n4)(3n), then at the point of equality you are multiplying by two negative numbers, which is the same as multilying by a positive number, so the direction of the < sign remains the same. The inequality becomes
3n(n4) <or= (n1)(3n4)
At this point you solve it the same as the equation, and get
12n <= 7n +4
5n <= 4
Now divide by the negative number 5 and reverse the direction of the > to get
n =or> 4/5
Check,
If n=1, n > 4/5 and
0/3 < 3/1 = 3
and the inequality is obeyed 
let's pretend this is an equation.
we would have
(n1)/(3n) = (n4)(3n4)
crossmultiplying, which is actually multiplying both sides by 3n(3n4) gives us
3n^2  7n + 4 = 3n^2  12n
n = 4/5
we also have critical values of n=0 and n= 4/3, these values make our expression undefined
BUT WE DID NOT HAVE AN EQUATION, but let's mark n = 4/5 or .8 , n = 0 and n = 4/3
on a number line
this divides our number line into 4 sections:
1 n < 4/5
2 n between 4/5 and zero
3 n between 0 and 4/3, and
4 n > 4/3
So let’s pick any arbitrary points in these regions,
for 1 pick any arbitrary point to the left of .8
e.g. n=1
left side of original is 2/3 which is 2/3
right side is 5/7 which is 5/7
is 2/3 ≤ 5/7 ? YES, so all values in that region will work
for 2 . try a number to the right of 4/3 but less than 0 , say n = 1/2
LS = 1 , RS = 5.5
Is 1 <  5.5 ? NO, so no values in that region will work
RS = 1/11 , is 4/15
Continue with a value in region in section 3 and 4 will show that
For 3 it does not work, but for 4 it does.
So……..
n ≤ 4/5 or 0 < n < 4/3
These type of questions are tougher that they appear. The problem is that
by multiplying by those denominators originally we have to consider if they are positive or negative, which would reverse the inequality sign 
1. n < 4/5
2. n between 4/5 and 0
3. n between 0 and 4/3
4. n > 4/3
So let's pick any arbitrary value in these sections
for 1. pick n = 1
LS = 2/3, RS = 5/7
is 2/3 < 5/7 YES, so all values in that section will work
section 2. let n = 1/2
LS = 1, RS = 5.5
is 1 < 5.5, NO, so no values in that region will work
I also tested section 3 and 4 and found
n ≤  4/5 OR 0 < n < 4/3 
I forgot about the critical value issue, and so will assume Reiny's answer is correct. I happened to test my answer at one n value that happened to be OK.