13

While working on this related inequality, I found the following inequality about right triangles. Let $a^2 + b^2 = 1$ be the two perpendicular sides of a right triangle then,

$$ 1 \le ab^{b^2/a^2} + ba^{a^2/b^2} < 1 + \frac{7}{180} $$

I found this inequality interesting because it is pretty tight in the sense that the upper bound is less than $3.89\%$ higher than the lower bound. Wolfram Alpha gives the upper bound as $1.03888$ and a Monte Carlo simulation gives $1.03888238314$ at $a = 0.18733386311638833$ and $b= 0.9822963013927571$. Hence I have used $1+\frac{7}{180} = 1.03888888...$ as an elegant estimate for the upper bound.

Question: The above upper bound is using brute force methods of calculus, What is the best upper bound that can be obtained using Olympiad level tools and standard well known inequalities?

Update 1: The lower bound follows immediately from Young's inequality. Let $a^2+b^2 = c^2$. Take $p = \frac{c^2}{b^2}$ and $q = \frac{c^2}{a^2}$ in $(3)$ of Young's inequality. Then, Young's inequality says that

$$ \frac{b^2}{c^2}a^{c^2/b^2} + \frac{a^2}{c^2}b^{c^2/a^2} = \frac{b^2}{c^2}a^{1+a^2/b^2} + \frac{a^2}{c^2}b^{1+b^2/a^2} = \frac{b^2}{c^2}a^{1+a^2/b^2} + \frac{a^2}{c^2}b^{1+b^2/a^2}\ge ab $$

or equivalently, $\displaystyle ab^{b^2/a^2} + ba^{a^2/b^2} \ge c^2$. Scaling the right triangle to a unit circle, we have $a^2+b^2=c^2 =1$ and the lower bound follows.

Nilotpal Sinha
  • 19,424
  • 1
    I'm not an Olympian; would applying Newton's method near the two outside zeros of the analytical derivative be consider the same order of "bruteness" as what you describe? I suppose so since I've used Wolfram Alpha to find the derivative, but presumably others could do that on paper. – uhoh Apr 23 '23 at 07:29
  • 2
    @uhoh Well all Olympiad inequalities can easily be solved by calculus and numerical analysis methods, although the computations could get tedious. Monte Carlo simulation with which I got the optimal values of $a,b$ achieves the same. But the reason I specifically asked for the best possible upper bound estimates using Olympiad method to see if it has an elegant solution something similar in split to this problem on MSE: https://math.stackexchange.com/questions/1410230/proving-that-e-pi-pie-lt-1-without-using-a-calculator – Nilotpal Sinha Apr 23 '23 at 07:41
  • 1
    @uhoh. I had an explicit solution using one single iteration of Newton method. This leads to a value of $1.03875$. I deleted it after the OP's comment. – Claude Leibovici Apr 23 '23 at 09:11
  • 1
    I am $99%$ certain the answer is no. The established literature for Olympiad problems is mostly about homogeneous inequalities and there is just not enough elementary methods non-homogeneous stuff like this with radicals in the exponent. – dezdichado Apr 23 '23 at 18:54
  • @NilotpalSinha Yes, that is nice. – River Li Apr 23 '23 at 22:25
  • @NilotpalSinha For the upper bound, I have a complicated analytic proof using bounds for functions in the expression. Do you have a proof? – River Li Apr 24 '23 at 03:14
  • @RiverLi I have a proof using calculus (equating the first derivative to zero) similar to the deleted answer below, but nothing without calculus. Does your method give an explicit closed from estimate of the upper bound (even if it is not the optimal bound).? – Nilotpal Sinha Apr 24 '23 at 03:17
  • @NilotpalSinha Equating the first derivative to zero, you don't have a closed form solution of the equation, right? I don't have a closed form of the upper bound. I mean to prove the inequality by using bounds such as $\ln x \le - \frac{268x + 3896}{10000x + 815}$ for all $x\in (0, 1/2]$. – River Li Apr 24 '23 at 03:58
  • @RiverLi Right, equating the first derivative to zero only gives a numerical approximation and not a closed form of the upper bound. Why don't you post your solution as a answer – Nilotpal Sinha Apr 24 '23 at 04:43
  • 2
    @NilotpalSinha Because it is not a proof using Olympiad tools and is complicated. If no proof in some days, I will post it. – River Li Apr 24 '23 at 06:32
  • 1
    When I see $a^2+b^2=1$, my first reflex is to say that $a=\sin(t)$ and $b=\cos(t)$. I find it particularly strange that nobody has thought of this yet. – Dominique Apr 26 '23 at 09:30

2 Answers2

3

If $$f(a)=a \left(1-a^2\right)^{\frac{1}{2} \left(\frac{1}{a^2}-1\right)}+a^{\frac{a^2}{1-a^2}} \sqrt{1-a^2}$$ compute the derivative to see that you need to solve for $a$ $$g(a)=2\,a^{\frac{a^2}{1-a^2}+3} \log (a)-\left(1-a^2\right)^{\frac{1}{2 a^2}+1} \log \left(1-a^2\right)=0$$ Using Taylor series around $a=0$ gives $$g(a)=\frac{a^2}{\sqrt{e}}+2 a^3 \log (a)+O\left(a^4\right)$$ So $$2 a \log (a)+\frac{1}{\sqrt{e}}=0 \quad \implies \quad a_0=\exp\left(W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right) \right)$$ where appears Lambert function. Taking into account the neglected terms, we know that this is an underestimate of the solution.

Numerically, this is $0.172660$ while the exact solution is $0.187334$.

Notice that $$f(a_0)=1.03875=1+\frac {31}{800}=1+\color{red}{\frac{279}{280}}\times \frac 7 {180}$$ which is not bad for a first shot.

Now, using the exact expression of $g(a)$, perform $\color{red}{\text{one single iteration}}$ of Newton method starting at $a_0$. The result is nasty but fully explicit and numerically $$a_1=0.189010$$ Halley method would give $a_1=0.187133$.

$$f(a_1)=1+\color{red}{\frac{3499263}{3500000}}\times \frac 7 {180}$$ and $\cdots\cdots$ we could do much better.

Claude Leibovici
  • 260,315
2

Some remarks using Bernoulli's inequality :

Denote :

$$f\left(x\right)=ab^{b^{2}/a^{2}}+ba^{a^{2}/b^{2}},a=\sqrt{x},b=\sqrt{1-x}$$

Then denote by $x_{max}\in(0,0.5)$ the maximum of $f(x)$ then using Bernoulli's inequality (tricky use) we have $x\in (x_{max}-\varepsilon,x_{max}+\varepsilon),\varepsilon\in[0,1/10000]$:

$$ab^{27}\left(1+\left(b-1\right)\left(-27+\frac{b^{2}}{a^{2}}\right)\right)-ab^{b^{2}/a^{2}}>0$$

Now we can use logarithm and derivative and note that it works not only for the coefficient $27$.

Hope it's useful for someone .

Edit :

I think we can do it without calculus but it seems very delicate :

We have :

Let be $a,b\in[0,0.25]$ as above and then $\exists c \in[0,\infty),\exists d \in[0,\infty)$ such that :

$$ba^{\frac{c}{d}}\left(1+\left(a^{\frac{1}{d}}-1\right)\left(-c+d\frac{a^{2}}{b^{2}}\right)\right)-ba^{a^{2}/b^{2}}\geq 0$$

It's the maximum we can get with Bernoulli's inequality accurate enought on small interval .It seems the very first step.

Go further :

For $x\in[0.0345,0.0355]$ and $a,b$ as above we have empiricaly :

$$ba^{a^{2}/b^{2}}+ab^{b^{2}/a^{2}}<ba^{\frac{176}{5000}}\left(1+\left(a^{\frac{1}{500}}-1\right)\left(-\frac{176}{10}+500\frac{a^{2}}{b^{2}}\right)\right)+ab^{27}\left(1+\left(b-1\right)\left(-27+\frac{b^{2}}{a^{2}}\right)\right)<1+7/180$$

One of the side is Bernoulli's with a twice use as above .

Then using simple bound we have $x\in[0.03506,0.035095]$

$$a^{\frac{176}{5000}}b\left(1+\left(\frac{3\left(a^{4}-1\right)}{896}\right)\left(-\frac{176}{10}+500\frac{a^{2}}{b^{2}}\right)\right)>ba^{\frac{176}{5000}}\left(1+\left(a^{\frac{1}{500}}-1\right)\left(-\frac{176}{10}+500\frac{a^{2}}{b^{2}}\right)\right)$$

Now there is the maxima but it seems too hard...

To be continued ...

Miss and Mister cassoulet char
  • 3,511