Don't they look equal?

$\large{99>2 \\ \Rightarrow 99 + \dfrac{1}{99} > 2 \\ \Rightarrow 99^{1000}\left(99 + \dfrac{1}{99}\right) > 2 \times 99^{1000} \\ \Rightarrow 99^{999} + 99^{1001} > 2 \times 99^{1000} \\ \Rightarrow 99^{2000} + 99^{999} + 99^{1001} + 1 > 99^{2000}+2\times 99^{1000} + 1 \\ \Rightarrow (99^{999}+1)(99^{1001}+1) > (99^{1000}+1)^2 \\ \Rightarrow \dfrac{99^{999}+1}{99^{1000}+1} > \dfrac{99^{1000}+1}{99^{1001}+1} \\ \Rightarrow \boxed{A>B}}$

$A = \frac{99^{999} + 1}{99^{1000} + 1}$

Multiply by 99 on Both Side.

$\implies 99A = \frac{99^{1000} + 99}{99^{1000} + 1}$ $\implies 99A = 1 + \frac{98}{99^{1000} + 1}$ $\implies 99A - 1 = \frac{98}{99^{1000} + 1}$

Similarly for B.

$99B - 1 = \frac{98}{99^{1001} + 1}$

Comparing RHS.

$\frac{98}{99^{1000} + 1} > \frac{98}{99^{1001} + 1}$ $\implies 99A - 1 > 99B - 1$ $\implies 99A > 99B$ $\implies \boxed{A > B}$

We will simply subtract B from A and check if it is greater or lesser than zero .......Now consider $A - B = \frac{1+99^{999}}{99^{1000}+1} - \frac{1+99^{1000}}{1+99^{1001}}$ $= \frac{99^{2000} + 99^{1001} + 99^{999} + 1 - (99^{2000} + 2.99^{1000} +1)}{k}$ where K is some large positive number ........CONSIDER THE NUMERATOR $= 99^{1001} + 99^{999} -2.99^{1000}$ this last expression is greater than zero (can be verified using AM GM inequality ) ...... hence $A - B > 0$ $A>B$

Lets make it quite simple Let it is (2^1+1)/(2^2+1)=A AND (2^2+1)/(2^3+1)=B and A= 3/5; B=5/9 Or A=0.6; B= 0.55 So A>B

$\frac{1}{A}=\frac{99^{1000}+1}{99^{999}+1}=99-\frac{98}{99^{999}+1}$ and $\frac{1}{B}=99-\frac{98}{99^{1000}+1}$ . Now $\frac{1}{B}>\frac{1}{A}$ since we are subtracting less from 99, so that $A>B$ .

Both equations have a form of (99^n + 1) / (99^(n+1) + 1) . Using concepts of Limit from Calculus we know that for n >= 1 the greater the n, the closer it's value to zero. So A is definitely greater.

Let:

$f(x) = \displaystyle\frac{{{{99}^x} + 1}}{{{{99}^{x + 1}} + 1}}$

Then:

$f'(x) = - \displaystyle\frac{{98 \times {{99}^x} \times \ln 99}}{{{{\left( {{{99}^{x + 1}} + 1} \right)}^2}}} < 0,\forall x \in \mathbb {R}$

Because $A = f(999)$ and $B=f(1000)$ and $f(x)$ is strictly decreasing, A must be greater than B.

Here is my unorthodox rationalization:

Step 1: First lets imagine that there were no +1s.

That would lead to:

$A_{2} = \frac{99^{999}} {99^{1000}}$

and

$B_{2} = \frac{99^{1000}} {99^{1001}}$

$B_{2} = A_{2} * \frac{99}{99} = A_{2} * 1 = A_{2}$

So, all of that is just thrown in there to mess you up and the real answer is in how the +1s modify the fractions.

Step 2: For fractions, what happens when you add a constant to the top and the bottom:

$\frac{1+1}{2+1} = \frac{2}{3} > \frac{1}{2}$

$\frac{1+2}{2+2} = \frac{3}{4} > \frac{2}{3}$

We can see here, that the larger the constant we add, the larger the resulting fraction.

Step 3:

$\frac{1}{99^{999} + 1} > \frac{1}{99^{1000} + 1} > \frac{1}{99^{1001} + 1}$

What does this mean? Adding 1 to the top and bottom of A and B has the effect of increasing the overall value of the fractions, but 1 is comparatively larger to 99^{999} than to 99^{1000} so it has a larger effect on A. Since we know that adding a larger value to the top and bottom of a fraction increases the overall value more than when adding a smaller value.

So, $\boxed{A > B}$

Just replace 99 by 2 and power 999 by 1, 1000 by 2, and so on and solve.

Take the reciprocal of B, which will have the same denominator as A, so you can compare them. You can see that 1/B is greater than A, it means that A is larger than B

B = 99×99^999 + 1/ 99×99^1000 + 1 taking 99 common and cancelling it we get B = (99^999 + 1/99) / (99^1000 + 1/99) which is smaller than A as at the place of 1 in A there is 1/99.

It seems to me that if you compare A with 1/B, it's obvious that 1/B > A and therefore that A > B. Am I making an obvious mistake?

I just thought of the graph of the exponetial function the further you go the faster you climb. So powers of N and N + 1 further up would be more spread out and their ratio (99^N / 99^N+1) would be smaller. With such large numbers adding 1 would be negligible so I looked past that.

If you cross multiply you should get 99^1001+99^999>2*99^1000.

This is how I did it :-)

99^999 < 99^1000 99^1001 >> 99^1000

Since a large number divided by an even larger number is less than 1, A has to be the largest. :-)

$99^{1000} - 99^{1001} > 99^{999} - 99^{1000}$ . If the denominator changes more, it is less. If the numerator changes more, it is more. In this case, the numerator changes more.

Adding the same number to the numerator and denominator of a proper fraction increases its value. $A=\frac{9^{1000} + 9}{9^{1001} + 9}$ . Since we are adding 8 to the numerator and denominator of B, its value increases.

If $A>B$ , then $\frac{A}{B}$ > 1; if $A<B$ , then $\frac{A}{B}$ < 1

$\frac{A}{B}$ = $\frac{(99^{999}+1)(99^{1001}+1)}{(99^{1000}+1)^2}$ = $\frac{99^{2000}+99^{1001}+99^{999}+1}{99^{2000}+2*99^{1000}+1}$ ; Therefore, $A>B$ iff $99^{1001}+99^{999} > 2*99^{1000}$

Therefore $A>B$ iff $99^2+1 > 2*99$ , which is true.

99^(1) is greater than 99^(-1)

First, let's multiply A by 99/99, so we'll have:

A = ((99^1000)+99)/((99^1001)+99)

Now, we can visualise a fraction composed of natural numbers to which we add equal amounts ((a+x)/(b+x)). We see that we have 3 cases for a/b: When a is greater than b, when they are equal, and when b is greater than a.

For case 1, we can imagine the most basic case, 1/2, and if we keep on adding the same amounts we get a consistently increasing value (2/3, 3/4, 4/5, etc), and consulting quite a few other cases, we arrive at the same conclusion.

In case 2, the fraction will always have a value of 1, no matter how much we add.

Case 3 is an analogy for case 1, because we have that the value of the fraction will constantly decrease, as every situation in this case is an inverse for one in case 1, so as the numbers get bigger, the inverse scenario takes place.

Knowing this, we can take A and B as (a+x)/(b+x) fractions, x being 99 in A and 1 in B. We can see that 99^1000 and 99^1001 are a and b respectively, so we can see it's a case 1 situation, and since we saw that the larger the amount added was, the larger the number, we can see that A will necessarily be greater than B.

I did it in the same way as Niranjan Ravikumar.

Simple, the more you elevate a number the grater it is. And now that we have the bigger number under the line, we know that is better to split the cake between 2 than between 10, so we look for the lower exponential number under the line, in this case 1000

We start with an inequality that isn't too hard to see why it's true. $99 > 2 - \frac{1}{99}$ Multiply both sides by 99 $99^{2} > 2(99) - 1$ Add 1 to both sides $1 + 99^{2} > 2(99)$ Multiply both sides by 99 to the 999 $99^{999} (1 + 99^{2}) > 2(99)(99^{999})$ $99^{999} + 99^{1001} > 2(99^{1000})$ Add 1 plus 99 to the 2000 to both sides $99^{999} + 99^{1001} + 99^{2000} + 1 > 2(99^{1000}) + 99^{2000} + 1$ Now we factor both sides.

The right side: Let U = 99 to the 1000 $2U + U^{2} + 1 = U^{2} + 2U + 1 = (U + 1)^{2} = (99^{1000} + 1)^{2}$

The left side: $99^{999} + 99^{1001} + 99^{2000} + 1$ $99^{2000} + 99^{999} + 99^{1001} + 1$ $99^{999}( 99^{1001} + 1) + (99^{1001} + 1)$ $(99^{1001} + 1)(99^{999} + 1)$

Now, rewrite the inequality with the factored expressions. $(99^{1001} + 1)(99^{999} + 1) > (99^{1000} + 1)^{2}$ Divide both sides by 1 plus 99 to the 1000 $\frac{(99^{1001} + 1)(99^{999} + 1)}{99^{1000} + 1} > 99^{1000} + 1$ Divide both sides by 1 plus 99 to the 1001 $\frac{99^{999} + 1}{99^{1000} + 1} > \frac{99^{1000} + 1}{99^{1001} + 1}$ Therefore, $\boxed{A>B}$

Easy. In this, we just need to compare A and B for us to come to any conclusion on what is greater. Take 99 common from the numerator and denominator in B which then leaves us with 99^999 + (1/99) in the numerator and 99^1000 + (1/99) in denominator which is certainly less than (99^999 + 1) /(99^1000+1)

if Y>X and N >1 then (X+1)/(Y+1) > (XN+1)/(YN+1) it is easy to prove this by simplification

if      Y>X   and N>1

==> Y(N-1) > X(N-1)

==> X+YN > Y+XN

==>(XYN+1)+X+YN>(XYN+1)+Y+XN

==>(X+1)(YN+1)>(Y+1)(XN+1)

==>(X+1)/(Y+1) > (XN+1)/(YN+1)

now substitute X = 99^999 , Y=99^1000 , N =99

I compared denominators to get the answer. The denominator of A is smaller in value than the denominator of B. We know that fractions with greater denominators are smaller(as in the case of B). So A is greater than B.

\frac{A}{B} will come greater than 1. So, A>B

just mutiply it and then subtract, easiest way

Let us substitute X for 99 and Y for the exponent 999. We can eliminate the +1 as it's common to all expressions.

Thus we get: A = X^Y / X ^ (Y+1) B = X ^ (Y+1) / X ^ (Y + 2)

(Y / Y+1) > ( (Y+1) / (Y+2) )

Therefor A > B.