On some inequality (upper bound) on a function of two variables

Question

There is a problem (of physical origin) which needs an analytical solution or a hint. Let us consider the following real-valued function of two variables

$y (t,a) = 4 \left(1 + \frac{t}{x(t,a)}\right)^{ - a - 1/2} \left(1 - \frac{2}{x(t,a)}\right)^{ 1/2} (x(t,a))^{-3/2} (z(t,a))^{1/4} \left(1 + \frac{t}{2}\right)^{ a},$

where $t > 0$, $0 < a < 1$ and

$x(t,a) = \frac{a-1}{2} t + \frac{3}{2}+\frac{1}{2}\sqrt{z(t,a)}$,

$z(t,a) = (1-a)^2 t^2+ 2(3-a)t +9.$

It is necessary to prove that

$$y(t,a) < 1$$

for all $t > 0$ and $0 < a < 1$. The numerical analysis supports this bound.

P.S. I apologize for too ``technical'' question. It looks that the inequality is valid also for $a = 1$ but it fails for $a > 1$.

Answer 1

Here is a human verifiable proof.

The quadratic equation for $x$ in terms of $a,t$ is $$ x^2-[(a-1)t+3]x+(2a-3)t=0 $$ Substituting $x=y+2$, we get $$ y^2+[1+(1-a)t]y=2+t=1+\frac{T-a}{1-a}\tag{*} $$ where $T=1+(1-a)t$ (I'm trying to mimic Mathematica explanation of proofs style). Thus $\frac{t}{2}+1=\frac{y(y+T)}{2}$.

Next, from the equation (*), $$ t=\frac{(y-1)(y+2)}{1-(1-a)y}\,, $$ so $$ 1+\frac tx=\frac{ay}{1-(1-a)y} $$ Finally, from the OP equation for $x$, we get $$ \sqrt z=2x-(a-1)t-3=2y+T\,. $$ Plugging all that nonsense in, we see that we need to bound from above (by $1$ or better) the expression $$ 4\left[\frac{(y+T)(1-(1-a)y)}{2a}\right]^a \left[\frac{(1-(1-a)y)}{ay}\right]^{1/2}\times \\ \left[\frac{y}{y+2}\right]^{1/2} \frac 1{(y+2)^{3/2}}(2y+T)^{1/2} $$ Now, from $(*)$, $$ (1-a)y(y+T)=1+T-2a\,, $$ so the first bracket simplifies to $$ \left[\frac{y-1+2a}{2a}\right]^a=\left[1+\frac{y-1}{2a}\right]^a\le 1+\frac{y-1}2=\frac{y+1}2 $$ (Here I use Bernoulli in the same way as Iosif did). We can cancel $y^{1/2}$ and combine $y+2$'s, which leaves us with the factor $$ F(T)=\frac{1-(1-a)y}a(2y+T) $$ Here I'm going to use the symbolic differentiation tool: $$ \frac d{dT}\log F(T)=-\frac{(1-a)\dot y}{1-(1-a)y}+\frac{2\dot y+1}{2y+T}\,. $$ From $(*)$, we have $(1-a)(2y+T)\dot y=1-(1-a)y$, so the logarithmic derivative evaluates to $\frac{2\dot y}{2y+T}\le \frac{2\dot y}{2y+1}=\frac d{dT}\log(2y+1)$ (using $\dot y>0$ and equality for $T=y=1$, of course). Thus $F(T)\le 2y+1$ and we end up with proving that $$ 2(y+1)\sqrt{2y+1}\le(y+2)^2 $$ However $(y+2)^2-2(y+1)\sqrt{2y+1}=\sqrt 2^2+(y+1-\sqrt{2y+1})^2$, which, according to the principle we discussed is trustworthy.

I leave the improvement of the last inequality to $2(y+1)\sqrt{2y+1}\le \gamma(y+2)^2$ with explicit $\gamma<1$ to Mathematica. Let me see if it is capable of that. But, of course, I want to see the full output, not just the declaration that the statement is true.

What I did doesn't go beyond the AI capabilities (just some bunch of random substitutions and symbolic manipulations). However, until I see a printout like this, I'll remain utterly skeptical. The fact that one of the most useful Mathematica commands for a friend of mine who is a big fan of it is "Quit kernel" doesn't add to my trust.

Answer 2

$\newcommand{\ty}{\tilde y}$With $x:=x(t,a)$, $y:=y(t,a)$, and $z:=z(t,a)$, we have \begin{align*} y &= 4(1+t/x)^{-a-1/2}(1-2/x)^{1/2} x^{-3/2} z^{1/4}(1+t/2)^a \\ &=4\Big(\frac{1+t/2}{1+t/x}\Big)^a(1+t/x)^{-1/2}(1-2/x)^{1/2} x^{-3/2} z^{1/4} \\ &\le\ty:=4\Big(1-a+a\frac{1+t/2}{1+t/x}\Big)(1+t/x)^{-1/2}(1-2/x)^{1/2} x^{-3/2} z^{1/4}, \end{align*} by the convexity of $u^a$ in $a$ for each real $u>0$.

Since $\ty$ is an algebraic function of $(t,a)$, it can be proved purely algorithmically that for all $(t,a)\in(0,\infty)\times(0,1)$ we have $\ty<1$ and hence $y<1$. Mathematica takes about 6 sec for this proof, which is a very long time for a computer program. In about 17 sec, Mathematica can also prove that $\ty<9/10$ and hence $y<9/10$. Below is the image of the corresponding Mathematica notebook; click on the image below to enlarge it.

Answer 3

a human verifiable proof:

Let us prove that the sharp bound of $y$ is $\frac{\sqrt{3+2\sqrt 3}}{3}$.

Letting $x := x(t, a), z := z(t, a), y := y(t, a)$, we have $$y = 4 \left(\frac{1 + t/2}{1 + t/x}\right)^a\cdot \left(\frac{z}{(1 + t/x)^2}\right)^{1/4} (1 - 2/x)^{1/2} x^{-3/2}. \tag{1}$$

Using Bernoulli inequality $(1 + u)^r \le 1 + ru$ for all $r \in (0, 1]$ and all $u > -1$, we have $$y \le 4 \left(1 + \left(\frac{1 + t/2}{1 + t/x} - 1\right)a\right)\cdot \left(\frac{z}{(1 + t/x)^2}\right)^{1/4} (1 - 2/x)^{1/2} x^{-3/2} \tag{2}$$ and $$y^2 \le 16 \left(1 + \left(\frac{1 + t/2}{1 + t/x} - 1\right)a\right)^2\cdot \left(\frac{z}{(1 + t/x)^2}\right)^{1/2} (1 - 2/x) x^{-3}. \tag{3}$$

We first simplify the expression. We have $$a \in (0, 1) \quad \iff \quad a = \frac{1}{1 + s}, \quad s > 0. $$ Then, we have (the so-called Euler substitution) $$t > 0 \quad \iff \quad t = \frac{2(1+s)(3s+2 - 3u))}{u^2 - s^2}, \quad s < u < s + \frac23. $$ With these substitutions, we have $$z = \left(\frac{-3u^2 + (6s + 4)u - 3s^2}{u^2 - s^2}\right)^2$$ and $$x = \frac{6s + 2}{u + s}.$$ Then, (3) is written as $$y^2 \le {\frac { \left(-3u^2 + 6su + 4u - 3s^2 \right) \left( 5\,s+2 - u \right) ^{2}}{4 \left( 3\,s+1 \right) ^{3}}}. $$ The constraints are: $s > 0$ and $s < u < s + \frac23$.

We have $$\sup_{s> 0, ~ s < u < s + \frac23} {\frac { \left(-3u^2 + 6su + 4u - 3s^2 \right) \left( 5\,s+2 - u \right) ^{2}}{4 \left( 3\,s+1 \right) ^{3}}} = \frac{3 + 2\sqrt 3}{9}. \tag{4}$$ (The proof is given at the end.)

Thus, we have $$y \le \frac{\sqrt{3 + 2\sqrt 3}}{3}.$$

On the other hand, in (1), letting $t = 3 + 3\sqrt 3, ~ a = 1$, we have $y = \frac{\sqrt{3 + 2\sqrt 3}}{3}$. Thus, $\frac{\sqrt{3 + 2\sqrt 3}}{3}$ is a sharp bound.

We are done.

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Proof of (4):

Fixed $s > 0$, let $$f(u) := {\frac { \left(-3u^2 + 6su + 4u - 3s^2 \right) \left( 5\,s+2 - u \right) ^{2}}{4 \left( 3\,s+1 \right) ^{3}}}.$$ The maximum of $f(u)$ is attained at $u_0 = 2\,s+1- \frac13\,\sqrt {9\,{s}^{2}+12\,s+3}$. Let $g(s) := f(u_0)$. It is easy to prove that $g'(s) < 0$ for all $s \ge 0$. Thus, $g(s) \le g(0) = \frac{3 + 2\sqrt 3}{9}$.

We are done.