Hey Veeky Forums, this is a problem I found on another forum which stumped me, here it is:
Let a,b be in (0,1), show that a^a+b^b >= a^b+b^a.
I ran a simulation and it appears to be true, I tried stuff like "assume that a>b" or assume that "a^a>b^b" but I didnt get anywhere.
This is not a homework problem... Even if it is I think it's pretty interesting
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This is the function f(x) = x^x , x = (0,1), in case you need it...
well if a=b that solves the "equals to" portion of the problem, and if a>b then a^a will always be greater than a^b and since its all exponentials b^a will never be able to pick up the slack from the a^b term to make up the difference, even if a is large
> if a>b then a^a will always be greater than a^b if a>b then a^a will always be greater than a^b
That is incorrect, if a>b then a^a will always be SMALLER than a^b, because since b is smaller, then a^b will be closer to 1 than a^a.
Also what do you mean by pick up the slack, could you elaborate on that please?
a>b => a^a>b^a and b^b>a^b => a^a+b^b > a^b+b^a
a>b does NOT imply b^b>a^b
If a>b, then from order from biggest to smallest is:
1. (a^b)
2. (a^a) & (b^b) (depends on the values of a&b)
3. (b^a)
i.e, the smaller the exponent, the bigger the term
I don't know about a formal proof, but here is an intuition about why this is correct, maybe it'll help you with the proof itself:
Let's assume that a is larger than b. If it helps you, imagine that a is much larger than b (e.g a is 0.9, b is 0.1).
a is closer to 1 than b, so it is less affected by exponents. No matter what power you raise it to, b will be more affected by that same power (because it is closer to 0). This also means that if you want to make you sum as big as possible, you shouldn't put too much "work" in making a larger, because it won't make it much bigger.
Instead, you should "put most of your money" on b. In terms of exponentiation, this means that you should give it the more effective power out of the two - i.e the smaller one (because smaller powers affect numbers more powerfully).
So you should give a the power a, because it won't be affected no matter what power you give it. And you should also give b the power b, because the smaller the power, the larger it will get (and like I said, b will get larger faster than a).
Hope this helps.
Ok I understand now (intuitively at least).
So in summary:
if (a>>b)
then b^b-b^a>=a^b-a^a is intuitive, because b is smaller and therefore more affected, hence the difference between the b's is bigger than the difference between the a's. Thanks!
Ran a quick simulation myself because I am autistic
dfrhs)
length(which(df$check == FALSE))
ggplot(df, aes(x = index))+
geom_point(aes(y = lhs), color = "firebrick", size = 1/50)+
geom_point(aes(y = rhs), color = "deepskyblue3", size = 1/50)
Just attach this image.
ok this problem seems fun!
i've tried working out with symmetry only using "half-angle", turns out (a+b)/2 and (a-b)/2 seem like good ideas but i'm stuck at the question
[math](a/b)^{\frac{a+b}{2}} \leq \frac{b^{\frac{a-b}{2}} - b^{\frac{b-a}{2}}}{a^{\frac{a-b}{2}} - a^{\frac{b-a}{2}}}[/math]?
even taking a look at the behavior of x-1/x (quotient of two of these on the right)
i need to go to bed, but you might want to try expressing one of the functions f(a,b) = a^a + b^b - a^b - b^a, or g(a,b) = (a^a + b^b)/(a^b + b^a) in terms of the symmetric variables s = a+b and d = a-b (maybe halved) and study the partial derivatives along d==0 axis...
i'll try again tomorrow, and hope you find before me so i can have that good feeling i get when i check my work and realise someone else's mind seem to agree with mine :)
have fun
For a fixed [math]b \in (0, 1)[/math], define [math]f \colon (0, 1) \to \mathbb{R}[/math] by [math]x \mapsto x^x + b^b - x^b - b^x[/math]. This is continuous, and [math]f'(x)=(\text{ln } x + 1)x^x -bx^{b-1} -b^x \text{ln }b[/math] has a root at [math]x=b[/math]. If you could check the behaviour of [math]f'[/math] on both sides of [math]b[/math], you would, very likely, find out it's where you get an extremum of the function. Then, you'd see that [math]f(b)=0[/math], and use this to get the desired result. Details are left for the reader.
I don't remember which inequality it was (could have been precisely this), but this is how I got the proof done.
That's what I would have done, but how do you show that b is a global minimum? You'd need to show that f'(x) has only one root at x = b (which it presumably does), but that would involve a lot more non-trivial work
How's the second derivative? If it has the same sign, or, equivalently, is non-zero everywhere on this interval, then you get the uniqueness out of that.
I wonder what the inequality was I proved, it wasn't this. and some fixing here, continuous -> differentiable.
take logs of both sides, subtract Rhs from Lhs, differentiate and investigate the first two derivatives
Saw this post yesterday, remembered I did this problem once in high school, but had completely forgotten since.
I'm doing a PhD in applied math now, so I asked my colleagues at lunch, including professors, and finally after 3 hours, a female PhD student came up with an elementary proof. Guess we're all brainlets lol.
Joking aside, this highlights quite well the difference between being good at math contest style problems and working in math.
I can't barely read the handwriting.
sorry, blame the thot that wrote it, also it's in French, but I'm too lazy to rewrite it, anyhow you can pretty much figure out everything from the math parts, she just used convexity and monotonicity, pretty elegant actually
"Revient à"
Can't you write some like everyone ?
again, it was a chick that wrote this, I was too lazy to rewrite/translate it, since from a mathematical point of view, everything is there, so problem solved
r8 h8 masturb8
I noticed after the fact that this proof also shows that the inequality is strict only when a = b
I meant attained, not strict
>convexity and monotonicity
...of [math]e^x[/math] and [math] \ln x[/math], respectively
the first ratio can be written
[math] \displaystyle \frac{a^a+b^b}{a^b+b^a}= \lambda e^{x_1}+(1- \lambda)e^{x_2}[/math]
where
[math] \displaystyle \lambda= \frac{b^a}{a^b+b^a}[/math]
and
[math] \displaystyle x_1= \left( \frac{a}{b} \right)^a=e^{a( \ln a- \ln b)}[/math]
and
[math] \displaystyle x_2= \left( \frac{b}{a} \right)^b=e^{b( \ln b- \ln a)}[/math]
then the result from convexity of [math]e^x[/math] is a bit simpler.
I tried to put in the most simple way I could:
Let [math] c = max{a,b} , d = min{a,b} [/math]
[math] \frac{c^c+d^d}{c^d + d^b} geq \frac{c^c+d^d}{d^c} = \left(\frac{c}{d}\right)^c + \frac{d^d}{d^c} [/math]
the first parcel is always greater than one.
If a==b, then, the quotient is equal to one.
Cheers
sorry, [math] c = max\{a,b\}, d = min\{a,b\} [/math]
and [math] \frac{c^c+d^d}{c^d+d^c} \geq \left(\frac{c}{d}\right)^c + \frac{d^d}{d^c} [/math]
But in your geq statement, c^d + d^c > d^c. Dividing by a bigger amount always yields a smaller number, not a bigger number.
Quelle école ?
not rigorous but i think this would solve the problem quite elegant
[math]
a^a-a^b+b^b-b^a\ge 0 \\
\Leftrightarrow \int_{b}^{a} {d\over dx} a^x + \int_{a}^{b} {d\over dx} b^x \,dx \ge 0 \\
\Leftrightarrow \int_{b}^{a} {d\over dx} a^x-b^x \,dx\ge 0 \\
\Leftarrow a^x \log a-b^x \log b\ge 0 \\
[/math]
the last part is easy if you assume [math]a>b[/math] and realize [math]x>0[/math]
ok turns out this doesnt work for values of b smaler than 0.03
for example 0.8^7 log 0.8-0.01^7 log 0.01 is negative