Finding the ratio of the volumes of a cube and a sphere

[Karol]

$$1 + x^\frac{3}{2} \leq (1+ x)$$
It's true if x<1, but why is that?
You assumed $~s_1>s_2$ but it needn't always be so

 

[Karol]

I ask again, what does it mean

suppose without loss of generality that $s_1\geq s_2$

That's why x<1 and the rest is true.
What if the sphere's area s₂>s₁?

 

[PeterDonis]

You assumed $s_1>s_2$ but it needn't always be so

The formula is symmetric in $s_1$ and $s_2$ (i.e., it's the same if you swap the two), so the case $s_2 > s_1$ is obviously the same as the case $s_1 > s_2$ (since the former is just the latter with $s_1$ and $s_2$ swapped).

 

[Karol]

You are right, in our case only the sphere's formula g(x) is used for both the cube and the sphere:
$$g(s_1) + g(s_2) \leq g(s_1 + s_2) = g(c)$$
So we can interchange between s₁ and s₂
But what if i don't want to interchange, s₂>s₁ thus x>1?

 

[StoneTemplePython]

You are right, in our case only the sphere's formula g(x) is used for both the cube and the sphere:
$$g(s_1) + g(s_2) \leq g(s_1 + s_2) = g(c)$$
So we can interchange between s₁ and s₂
But what if i don't want to interchange, s₂>s₁ thus x>1?

The idea is that the relation is invariant to permutations of the orderings (in terms of size) of $s_i$. In general if you don't like this sort of argument and you have $n$ terms you can run the argument over $n!$ different special cases. In your problem $n=2$ and $2! =2$ which is easy enough. For bigger problems, well, watch out, because $n!$ grows outrageously fast. Hence you should look for a way to break the underlying symmetry -- that is the reason for the Without Loss of Generality argument.

In the event don't want to use such an approach you can instead do it by cases:

case 1: if $s_1 \geq s_2$
then multiply each side by
$\big(\frac{1}{s_1}\big)^\frac{3}{2}$

and get

$1 + x^\frac{3}{2}= 1 + \big(\frac{s_2}{s_1}\big)^\frac{3}{2} \leq (1 + \frac{s_2}{s_1})^\frac{3}{2} = (1 + x)^\frac{3}{2}$

(now finish it off)
case 2: if $s_2 \gt s_1$
then multiply each side by
$\big(\frac{1}{s_2}\big)^\frac{3}{2}$

and get

$1 + x^\frac{3}{2}= 1 + \big(\frac{s_1}{s_2}\big)^\frac{3}{2} \leq (1 + \frac{s_1}{s_2})^\frac{3}{2} = (1 + x)^\frac{3}{2}$

(now finish it off -- but notice how the argument is identical to case 1, you just need to prove $1 + x^\frac{3}{2} \leq (1 + x)^\frac{3}{2}$ ... so why bother with doing a second case? )

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edit:
as for your other question:

If we assume $s_1\geq s_2$ then $0 \leq x \leq 1$, is it the reason?

yes that is correct. In general isolating variables to be between 0 and 1 is useful and gives some nice structure.