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

[StoneTemplePython]

I have to prove that $~c^d+e^d\leq(c+e)^d~$
Newton's binomial's: $(a+b)^n=C^0_n a^n+C^1_n a^{n-1}b+...+C^n_n b^n$
$$\rightarrow~a^n+b^n\leq(a+b)^n$$
Because $~a^n+b^n~$ are only part of the binomial.
But in this method you can't find V_min

note that the inequality holds for real non-negative $c$ and $e$ and for any $d \geq 1$

Your instincts to expand the binomial are good -- but note that $d$ is not positive integer here. There are some generalizations of binomial expansions with non-integer powers but they're kind of ugly.

I'd suggest just squaring each side and simplifying the algebra. (A fancier approach is to re-write it as $(c^d+e^d)^\frac{1}{d} \leq c+e$ which is true by direct application of Minkowki Inequality, but just squaring each side and simplifying gets you there.)

I'll post the minimization approach a bit later today. Your problem was seemingly Taylor made for Lagrange Multipliers. The approach I'll post doesn't use them (though if you know where to look you can see them lurking nearby) and just uses linearity and convexity. The approach is a touch longer but it has a very nice visual component to it which I think could be very beneficial in understanding why the solution you found is correct.

 

[StoneTemplePython]

Minimization problem

main goal: practice using, drawing, and visualizing tangent lines and how they interact with convexity

What's the typical approach for minimizing a convex function? Setting derivative equal to zero. Suppose you have two functions $f$ and $g$ as in above, and your constraint is a linear combination of them. Why not set the derivative of each of them equal to zero -- which means you could set the derivative of one function equal to the other? (Technical nit -- setting both derivatives equal to zero is too strict a goal here and would lead to violating the constraint, but setting the derivatives equal to each other gives you more flexibility, and as will be shown, ultimately works.) I.e. if we pursue this, we find that there are these optimal points $s_1^*$ and $s_2^*$ given by working through $f'(s_1) = g'(s_2)$.

At these respective points if you evaluate the derivative, you get a tangent line for each function. The slope of said line is the same (because the derivatives are the same).

So what's the equation of each line (that doesn't necessarily go through the origin)?

These tangent lines at our optimal points are now completely determined and have very familiar formulas of

$A_f^*(x) = mx + b_f$
$A_g^*(x) = mx + b_g$

(i.e. they are technically affine because in general the lines don't go through the origin -- draw a picture to confirm!)

Again, the slopes of these tangent lines are equal -- both are equal to some fixed value of $m$ which you can calculate. We can now, with some care, harness linearity:

$A_f^*(s_1) + A_g^*(s_2) = \big(ms_1 + b_f\big) + \big(m s_2 + b_g\big) = m(s_1 + s_2) + b_f + b_g = m(c) + b_f + b_g$

which is valid, and constant, for any legal choice of $s_1$ and $s_2$
- - - -
The magic is that since $f$ and $g$ are differentiable convex functions over positive numbers, we know that the tangent lines are linear lower bounds of the function elsewhere. This is a very nice property for convex functions. (Note that if the functions were not differentiable, but were convex, we'd still have linear lower bounds, they just wouldn't be tangent lines per se.)

I.e. at any and every point on the function, the given tangent lines go through the point of tangency, and *everything* else given by the function (over our domain) is above the tangent line -- if you're not familiar with this, I strongly recommend drawing the picture here for things like $y = x^2$ and $y= e^x$ and sketching a few tangent lines -- the visualization is very helpful. (Technical nit: the above is the story for strict convexity, for non-strict convexity I should adjust the wording slightly to say that tangent line goes through the point of tangency, and nothing else on the function is below tangent line)

- - - -
How do I know that $s_1^*$ and $s_2^*$ in fact give a minimum for our problem? Well consider the following, for a contradiction:

Suppose there is a better solution than the one we've found and this better solution is given by $\bar{s_1}, \bar{s_2}$, where as always both are real non-negative, and $\bar{s_1} + \bar{s_2} = c$, but $\bar{s_1} \neq s_1^*$ and $\bar{s_2} \neq s_2^*$, then if we examine the tangent line as a lower bound, we see:

$f(\bar{s_1}) \geq A_f^*(\bar{s_1})$

and also

$g(\bar{s_2}) \geq A_g^*(\bar{s_2})$

because of the linear lower bound property of tangent lines of convex functions. But, if we put everything together:

$f(\bar{s_1}) + g(\bar{s_2}) \geq A_f^*(\bar{s_1}) + g(\bar{s_2}) \geq A_f^*(\bar{s_1}) + A_g^*(\bar{s_2}) = A_f^*(s_1^*) + A_g^*(s_2^*) = f(s_1^*) + g(s_2^*)$

which contradicts the fact that we've found a better solution.

(please sketch this out for a very immediate, visual interpretation of these inequalities -- the idea is that tangent line given by $A_f^*$ is equal to the $f(s_1^*)$, by design, and $A_g^*$ does so at $g(s_2^*)$ but in general these tangent lines are each lower bounds for everything else. If you sketch this out it's a very, very nice result.)

Hence there can be no better solution than that given by using $s_1^*$ and $s_2^*$. In fact this is strictly the best solution, which we can show by tightening the argument slightly via use of strict convexity, though I don't think it is needed for the exercise. Another nit: framing this as a contradiction isn't technically needed but it feels a lot like an exchange argument, which I like, so I went with it.

This completes the minimization problem. What I was hoping you'd get is some kind of visualization for what's going on in the minimization problem, and I think you'll get it if you sketch out the functions and some relevant tangent lines.
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The approach here is awfully close to that in Lagrange Multipliers, but I tried to evade partial derivatives and said multiplier in favor of single variable derivatives and tangent lines -- which we can combine via linearity and bound the results because of convexity.

it may be worthwhile to confirm that $f$ and $g$ are (strictly) convex over positive numbers. The easy way in this case is to differentiate each function twice and show the second derivatives are always positive.

 

[Karol]

I'd suggest just squaring each side and simplifying the algebra.

I have to prove $~~c^d+e^d\leq(c+e)^d~$:
$$(c^d+e^d)^2=c^{2d}+2c^de^d+e^{2d}=c^{2d}+2(ce)^d+e^{2d}\leq(c+e)^{2d}=(c^2+2ce+e^2)^d$$
I can't simplify no longer on the right side since d isn't a positive integer

 

[StoneTemplePython]

I have to prove $~~c^d+e^d\leq(c+e)^d~$:
$$(c^d+e^d)^2=c^{2d}+2c^de^d+e^{2d}=c^{2d}+2(ce)^d+e^{2d}\leq(c+e)^{2d}=(c^2+2ce+e^2)^d$$
I can't simplify no longer on the right side since d isn't a positive integer

If you're going to take the square and simplify approach, then it's better to make this less abstract -- in this problem $d :=\frac{3}{2}$ i.e. you are proving

$g(s_1) + g(s_2) = \Big(\frac{s_1^\frac{3}{2}}{6 \pi^\frac{1}{2}}\Big) + \Big(\frac{s_2^\frac{3}{2}}{6 \pi^\frac{1}{2}}\Big) \leq \Big(\frac{(s_1 + s_2)^\frac{3}{2}}{6 \pi^\frac{1}{2}}\Big) = g(s_1 + s_2)$

you can always re-scale by positive numbers without changing the inequality, so this reduces your problem to proving

$s_1^\frac{3}{2} +s_2^\frac{3}{2} \leq (s_1 + s_2)^\frac{3}{2}$
noting that both sides must be positive (so squaring is invertible and preserves ordering)
if you square both sides, you get

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

which simplifies to
$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2$

if either $s_1 = 0$ or $s_2 =0$ then the LHS =0 and the RHS = 0 which verifies the above relationship.

Alternatively they could both be positive, so to finish it off, assuming both $s_1 \gt 0$ and $s_2 \gt 0$, simplify the relationship and show the final result...

 

[Ray Vickson]

Homework Statement

View attachment 222596

Homework Equations

Volue of a sphere: $~\displaystyle V=\frac{4}{3}\pi r^3$
Area of a sphere: $~\displaystyle A=4\pi r^2$
Minimum/Maximum occurs when the first derivative=0

The Attempt at a Solution

The fixed area is k, the edge is a:
$$6a+4\pi r^2=k~\rightarrow~a=\frac{k-4\pi r^2}{6}$$
$$V=a^3+\frac{4}{4}\pi r^3=\left[ \frac{k-4\pi r^2}{6} \right]^3+\frac{4}{4}\pi r^3$$
$$V'=\frac{3}{216}(k-4\pi r^2)^2(-8)\pi r+4\pi r^2$$
$$V'=0:~~4\pi r-\frac{8}{72}(k-4\pi r^2)^2\pi=0~\rightarrow~r=\frac{3k-2}{12\pi}$$
$$\frac{a}{r}=\frac{(k-4\pi r^2)12\pi}{6(3k-2)}=\frac{\pi(k-4\pi r^2}{3k-2}$$
The k doesn't cancel

Now that the problem has been solved, let me offer a simple alternative solution. You want to solve
$$\begin{array}{cc}
\max / \min & \rm{Vol} = a^3 + \frac{4}{3} \pi r^3 \\
\rm{s.t.} & A = 6 a^2 + 4 \pi r^2 = A_0\\
& a \geq 0, r \geq 0
\end{array}
$$
for some fixed value $A_0 > 0$. Since the cubic and spherical areas and volumes scale in the same way, we might as well let $A_0 = 1$, which we now do.

Changing variables to $x,y$, with $a = (1/\sqrt{6}) \: x$ and $r = (1/\sqrt{4 \pi}) \: y$ gives the constraint as $x^2+y^2 = 1$ and the volume as
$$\rm{Vol} = V(x,y) = \frac{\sqrt{6}}{36} x^3 + \frac{1}{6 \sqrt{\pi}} y^3.$$
Going to polar coordinates $x = \cos(t), y = \sin(t)$ gives the volume as ${\rm{Vol} }= v(t)$:
$$v(t) = \frac{\sqrt{6}}{36} \cos^3(t) + \frac{1}{6 \sqrt{\pi}} \sin^3(t)$$.
Here is a plot of $v(t)$ for $0 \leq t \leq \pi/2$:

Finding closed-form expressions for the values of $t$ that set $dv/dt = 0$ is straightforward.

 

[Orodruin]

Since the areas and volumes scale in the same way, we might as well let A0=1A0=1A_0 = 1

Just to clarify this, I believe you are saying that the volume of the sphere scales the same way as that of the cube and the same thing for the areas, but it may be misread as saying that the volume and area scale in the same way, which they do not. Either way, I don't see a real point in setting $A_0 = 1$ as it is anyway just a scale that factors out of everything (i.e., set $a = x\sqrt{A_0/6}$ etc. and you still get $x^2 + y^2 = 1$).

 

[Ray Vickson]

Just to clarify this, I believe you are saying that the volume of the sphere scales the same way as that of the cube and the same thing for the areas, but it may be misread as saying that the volume and area scale in the same way, which they do not. Either way, I don't see a real point in setting $A_0 = 1$ as it is anyway just a scale that factors out of everything (i.e., set $a = x\sqrt{A_0/6}$ etc. and you still get $x^2 + y^2 = 1$).

I meant that when we replace $V_0$ by $k V_0$ we need only scale both $a$ and $r$ by $k^{1/3}$, so both areas scale as $k^{2/3}$ and both volumes scale as $k$. Optimality is preserved.

 

[Karol]

$s_1^3 +s_2^3 + 2(s_1 s_2)^\frac{3}{2} \leq (s_1 + s_2)^3 = s_1^3 + 3s_1^2s_2 + 3 s_1 s_2^2 + s_2^3$
which simplifies to
$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2$

$$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2=3s_1s_2(s_1+s_2)$$
But $~(s_1 s_2)^\frac{3}{2}\geqslant s_1 s_2$
$$s_1 s_2\leq(s_1 s_2)^\frac{3}{2} \leq \frac{3}{2}s_1s_2(s_1+s_2)$$
$$1\leq\frac{3}{2}(s_1+s_2)$$
If s₁ and s₂ > 1 it is shurley true. it might be true in some cases if these are <1. do they?

 

[StoneTemplePython]

$$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2=3s_1s_2(s_1+s_2)$$
But $~(s_1 s_2)^\frac{3}{2}\geqslant s_1 s_2$
$$s_1 s_2\leq(s_1 s_2)^\frac{3}{2} \leq \frac{3}{2}s_1s_2(s_1+s_2)$$
$$1\leq\frac{3}{2}(s_1+s_2)$$
If s₁ and s₂ > 1 it is shurley true. it might be true in some cases if these are <1. do they?

Try another approach. When you look at

$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2$

Notice that everything is positive on each side so you can always divide out the positive items without changing the inequality.

What happens if you divide each side by $2(s_1 s_2)^\frac{3}{2}$? Note you can verify that resulting inequality by inspection, but a more slick approach is to lower bound the resulting right hand side via $GM \leq AM$ -- something we've certainly used in these threads before. I think you can get there.

- - - -
edit:
alternatively, consider running with this factorization

$$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2=3s_1s_2(s_1+s_2)$$

again, we've assumed $s_1 \gt 0$ and $s_2 \gt 0$ so why not divide both sides by $(s_1 s_2)$ which you've isolate on the RHS. This gives you

$2(s_1 s_2)^\frac{1}{2} = 2 GM \leq 2 AM \lt 3(s_1+s_2)$

perhaps you can interpret this and fill in the blank for what the resulting value of 2 times Arithmetic mean is?

 

[Karol]

$$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2~\rightarrow~1\leq\frac{3}{2}\left[ \frac{s_1^2s_2}{\sqrt{(s_1s_2)}}+\frac{s_1s_2^2}{\sqrt{(s_1s_2)}} \right]$$
$$1\leq\frac{3}{2}\left[ \sqrt{\frac{s_1}{s_2}}+\sqrt{\frac{s_2}{s_1}} \right]$$
It resembles $~\displaystyle \frac{1}{x}+\frac{x}{1}\geq1$
The first $~GM \leq AM~$ method:
$$GM \leq AM~\rightarrow~\frac{s_1^2s_2+s_1s_2^2}{2}\geq\sqrt{(s_1^2s_2)(s_1s_2^2)}=\sqrt{(s_1s_2)^3}=(s_1s_2)^{3/2}$$
$$3s_1^2s_2 + 3 s_1 s_2^2=3(s_1^2s_2+s_1s_2^2)\geq3(s_1s_2)^{3/2}\geq2(s_1s_2)^{3/2}$$
And the second approach:
$$GM \leq AM~\rightarrow~2\sqrt{s_1s_2}\leq2\frac{s_1+s_2}{2}=s_1+s_2\leq2(s_1s_2)\leq3(s_1s_2)$$
Which makes true:
$$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2=3s_1s_2(s_1+s_2)\rightarrow~
2(s_1 s_2)^\frac{1}{2}\leq3(s_1s_2)$$

 

[StoneTemplePython]

$$2(s_1 s_2)^\frac{3}{2} \leq 3s_1^2s_2 + 3 s_1 s_2^2~\rightarrow~1\leq\frac{3}{2}\left[ \frac{s_1^2s_2}{\sqrt{(s_1s_2)}}+\frac{s_1s_2^2}{\sqrt{(s_1s_2)}} \right]$$

The right hand side isn't right, but maybe that's a typo? If you're going to show it this way the RHS should say

$\frac{3}{2}\left[ \frac{s_1^2s_2}{s_1 s_2 \sqrt{(s_1s_2)}}+\frac{s_1s_2^2}{s_1 s_2 \sqrt{(s_1s_2)}} \right]$

Your very next line is correct.

$$1\leq\frac{3}{2}\left[ \sqrt{\frac{s_1}{s_2}}+\sqrt{\frac{s_2}{s_1}} \right]$$
It resembles $~\displaystyle \frac{1}{x}+\frac{x}{1}\geq1$

Yes. And that is what you can verify by inspection if you wanted -- i.e. if $0 \lt x \lt 1$, then you have the first term greater than 1 and second term is positive, so sum is $\gt 1$. If $1 \leq x$ then second term is at least one, and first term is positive so the sum is again $\gt 1$. From there, of course, you multiply by 1.5 -- the product of any number $\gt 1$ and $1.5$ must be $\gt 1$.

From $GM \leq AM$ standpoint, it resembles

$\frac{1}{2}\big(\frac{1}{x}+\frac{x}{1}\big) \geq 1$

because by $GM \leq AM$, you have

$\frac{1}{2}\big(\frac{1}{x}+\frac{x}{1}\big) \geq \big(\frac{x}{1}\frac{1}{x}\big)^\frac{1}{2} = \big(1 \big)^\frac{1}{2} = 1$

now all you need to do is note that -- again since everything is positive --

$3\Big(\frac{1}{2}\big(\frac{1}{x}+\frac{x}{1}\big)\Big) \gt \frac{1}{2}\big(\frac{1}{x}+\frac{x}{1}\big) \geq 1$

and define $x:= \sqrt{\frac{s_1}{s_2}}$ and you are done.
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The first $~GM \leq AM~$ method:
$$GM \leq AM~\rightarrow~\frac{s_1^2s_2+s_1s_2^2}{2}\geq\sqrt{(s_1^2s_2)(s_1s_2^2)}=\sqrt{(s_1s_2)^3}=(s_1s_2)^{3/2}$$

Yes.

$$3s_1^2s_2 + 3 s_1 s_2^2=3(s_1^2s_2+s_1s_2^2)\geq 3(s_1s_2)^{3/2}\geq2(s_1s_2)^{3/2}$$

Maybe you are skipping steps here? What happened to the $\frac{1}{2}$ on the LHS and how does the 3 flow though? If it were me, I'd be very deliberate and take the conclusion from your prior statement, i.e. that

$\frac{s_1^2s_2+s_1s_2^2}{2}\geq (s_1s_2)^{3/2}$

then step 1: multiply each side by 2

$s_1^2s_2+s_1s_2^2 \geq 2(s_1s_2)^{3/2}$

step 2: note that 3 of some positive number is bigger than that positive number, i.e.

$3(s_1^2s_2+s_1s_2^2) \gt s_1^2s_2+s_1s_2^2 \geq 2(s_1s_2)^{3/2}$

or if you prefer, the slightly looser, but true, statement that

$3(s_1^2s_2+s_1s_2^2) \geq s_1^2s_2+s_1s_2^2 \geq 2(s_1s_2)^{3/2}$

and you are done.

And the second approach:
$$GM \leq AM~\rightarrow~2\sqrt{s_1s_2}\leq2\frac{s_1+s_2}{2}=s_1+s_2\leq2(s_1s_2)\leq3(s_1s_2)$$

yikes -- some careless errors here -- missing plus signs (and I'm not sure what the introduction of $\leq2(s_1s_2)$ on the RHS was for?) This should read as

$GM \leq AM~\rightarrow~2\sqrt{s_1s_2}\leq2\frac{s_1+s_2}{2}=s_1+s_2 \leq3(s_1 + s_2)$which gives us what we want: $2\sqrt{s_1s_2}\leq 3(s_1 + s_2)$

but yes, once you have the above you simply multiply each side by $(s_1 s_2)$ -- which is positive, and hence doesn't change the inequality-- and you recover

$2(s_1 s_2)^\frac{3}{2} =(s_1 s_2) 2\sqrt{s_1s_2}\leq (s_1 s_2)3(s_1 + s_2) = 3(s_1^2 s_2) + 3(s_1 s_2^2)$

- - - -
it's good practice to play around with inequalities like this. And hopefully you can spot things like

$\frac{1}{x}+\frac{x}{1}\geq1$

lurking amongst things you want to prove in the future. It's a quite nice result.

 

[Karol]

Thus, for the maximization problem, for real non-negative numbers -- call them $s_1$, $s_2$, we have the below inequality

$f(s_1) + g(s_2) \leq g(s_1) + g(s_2) \leq g(s_1 + s_2) = g(c) = f(0) + g(c)$

After you have that proven, the maximization problem is done -- i.e. the Right Hand Side of the inequality tells you that you get the best possible 'score' when you allocate everything to $g$ which maps to sphere volume.

I am not sure i understand. i think i can satisfy with only:
$V_{general}=f(s_1) + g(s_2) \leq g(s_1) + g(s_2)=g(s_1=0)+g(s_2=max)=0+g(s_2=max)$
Since $s_1+s_2=A_0~$ and when $s_1=0~$, meaning there is no cube, then $s_2=max~$ is only the aphere

 

[StoneTemplePython]

I am not sure i understand. i think i can satisfy with only:
$V_{general}=f(s_1) + g(s_2) \leq g(s_1) + g(s_2)=g(s_1=0)+g(s_2=max)=0+g(s_2=max)$
Since $s_1+s_2=A_0~$ and when $s_1=0~$, meaning there is no cube, then $s_2=max~$ is only the aphere

Welcome back!

In the interim I discarded my train of thought here... it was something clever with linearity and convexity but I'll have to revert a bit later.

 

[StoneTemplePython]

I am not sure i understand. i think i can satisfy with only:
$V_{general}=f(s_1) + g(s_2) \leq g(s_1) + g(s_2)=g(s_1=0)+g(s_2=max)=0+g(s_2=max)$
Since $s_1+s_2=A_0~$ and when $s_1=0~$, meaning there is no cube, then $s_2=max~$ is only the aphere

it's coming back to me... the idea is

$V_{general}=f(s_1) + g(s_2) \leq g(s_1) + g(s_2) \leq g(s_1 + s_2) = g(c) = 0 + g(c) = f(0) + g(c)$

where I used some constant $c \gt 0$ but, equivalently, to bridge the notation $c = A_0 = s_1 + s_2$
- - - -
the key point, though is, do you believe

$g(s_1) + g(s_2) \leq g(s_1 + s_2)$

i.e. it in general is an inequality, not an equality. Why is this true?

- - - -
edit:

for avoidance of doubt, I'd say:

$g(s_1) + g(s_2)\neq g(s_1=0)+g(s_2=max)=0+g(s_2=max)$

for arbitrary legal $s_1, s_2$ on the left hand side. It's possible you meant something else here notationally, but the thing to keep in mind is we want to show, not only that $g$ "beats" $f$ for maximizing, but also that $g$ attains its maximum at $c$ -- that's the purpose of the second inequality

$g(s_1) + g(s_2) \leq g(s_1 + s_2) = g(c)$

to be certain that we have a maximum when we allocate everything $g(c)$...

maybe it was too much symbol manipulation to get to this final inequality. A different but easy idea is: go back to post 39:

$s_1^\frac{3}{2} +s_2^\frac{3}{2} \leq (s_1 + s_2)^\frac{3}{2}$

suppose without loss of generality that $s_1\geq s_2$ (This means $s_1 \gt 0$ btw)

now multiply each side by $\big(\frac{1}{s_1}\big)^\frac{3}{2}$ . This should give you

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

this implies $0 \leq x \leq 1$. (Why?) You could apply Newton's binomial formula here though I don't think that helps with intuition. Before we did something similar and battered it with $\text{GM} \leq \text{AM}$.

A better, and succinct finish would be to note

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

why is that true?

 

[Karol]

suppose without loss of generality that $s_1\geq s_2$ (Why assume that?)
$1 + x^\frac{3}{2}= 1 + \big(\frac{s_2}{s_1}\big)^\frac{3}{2} \leq (1 + \frac{s_2}{s_1})^\frac{3}{2} = (1 + x)^\frac{3}{2}$
this implies $0 \leq x \leq 1$. (Why?)

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

 

[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.