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Posting from MSE since it was unanswered in MSE.

Let $0 \le x,y \le 1$ and $a$ be a real and let

$$ f(x,y,a) = xy^{y^a/x^a} + yx^{x^a/y^a} - x^a - y^a \tag 1 $$

For a fixed $a$, the graph of the maximum and the minimum value of $f(x,y,a)$ shown below.

enter image description here

The graph shows that for $a \ge 1$, there is a constant $C_a$ depending only in $a$ such that $xy^{y^a/x^a} + yx^{x^a/y^a} - y^a - x^a \le C_a$ and similarly for $a \le 2$, there is a constant $c_a$ such that $c_a \le xy^{y^a/x^a} + yx^{x^a/y^a} - y^a - x^a$. Also, $0\ <a < 2$ is the only interval in which $f(x,y,a)$ has non-zero maxima and minima for every $a$.

Some experimentally observed examples of inequalities belonging to this family are

$$ -\frac{1}{2} \le xy^{y/x} + yx^{x/y} - x - y < 0 \tag 2 $$

$$ 0 \le xy^{y^2/x^2} + yx^{x^2/y^2} - x^2 - y^2 \le \frac{1}{4} \tag 3 $$

$$ 0 \le xy^{y^4/x^4} + yx^{x^4/y^4} - x^4 - y^4 \le \frac{1}{2} \tag 4 $$

Question 1: Can we prove that for every real $a > 0$, there exists $c_a$ and $C_a$ such that

$$ c_a \le xy^{y^a/x^a} + yx^{x^a/y^a} - x^a - y^a \le C_a $$

Question 2: Can we express $c_a$ and $C_a$ in terms of $a$?

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    $\begingroup$ The lower bound when $a \leq 2$ is probably attained along the line $x = y$; if that is true than an elementary computation would give $c_a = 2\left[ \left( \frac2a\right)^{\frac2{a-2}} - \left( \frac2a \right)^{\frac{a}{a-2}} \right]$. $\endgroup$ May 9, 2023 at 14:21
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    $\begingroup$ The upper bound when $a \geq 1$ is probably attained when one of $x,y$ tends to zero. If this is true than a computation would show $C_a = \left(\frac1a\right)^{\frac1{a-1}} - \left(\frac1a\right)^{\frac{a}{a-1}}$. $\endgroup$ May 9, 2023 at 14:28
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    $\begingroup$ actually: do you have numerical simulations for $a \in [3,10)$? My previous comment is wrong. When $a$ is large, it seems that the maximum may also be attained on the diagonal. $\endgroup$ May 9, 2023 at 15:21

2 Answers 2

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As Iosif observed, you have $-2 \leq c_a \leq C_a \leq 2$ for any $a$.

Slightly improved upper bounds

When $a < 1$

The expression is symmetric in $x$ and $y$, so we can WLOG assume $y \geq x$. Let $\beta = y^a / x^a$ and $\beta^* = y^a / (y^a - x^a)$ its Holder conjugate. Then Young's inequality states

$$\tag{*} x y^\beta \leq \frac{x^{\beta^*}}{\beta^*} + \frac{y^{\beta^2}}{\beta} $$

and

$$\tag{**} x^{\beta^{-1}} y \leq \frac{x}{\beta} + \frac{y^{\beta^*}}{\beta^*} $$

Using that $a < 1 \leq \beta,\beta^*$ and that $x,y \leq 1$, we therefore conclude

$$ xy^\beta + y x^{\beta^{-1}} \leq (x^a + y^a) \left( \frac{1}{\beta} + \frac{1}{\beta^*} \right) = x^a + y^a $$

and hence $f(x,y,a) \leq 0$ for all $a < 1$. Taking the limit as $x,y \to 0$ for any fixed $a$ shows that this estimate is sharp. This proves that the best $C_a$ when $a < 1$ is $C_a = 0$.

When $a \geq 1$

Expressions (*) and (**) still hold in general, so the same argument proves that

$$ x y^\beta + y x^{\beta^{-1}} \leq x + y \implies f(x,y,a) \leq x - x^a + y - y^a $$

This expression can be optimized using single variable calculus to obtain the upper bound

$$ f(x,y,a) \leq 2 \left[ \left( \frac1a\right)^{\frac1{a-1}} - \left( \frac1a \right)^{\frac{a}{a-1}} \right] < 2 $$

(Note that this is a worse bound than was proposed in my comments to the OP.)

Sharp lower bounds

When $a \geq 2$

Again, since the expression is symmetric in $x,y$, we can assume WLOG that $y \leq x$; write $y = \lambda x$ for $\lambda \in (0,1]$. Our function

$$ f = x^{1+\lambda^a} \lambda^{\lambda^a} + x^{1 + \lambda^{-a}} \lambda - x^a (1 + \lambda^a) $$

Observe that

$$ (1+\lambda^a)^{-1} + (1 + \lambda^{-a})^{-1} = 1 $$

and so we can apply Young's inequality (writing the first two terms as $A^{1+ \lambda^a} / (1 + \lambda^a)$ and $B^{1+\lambda^{-a}} / (1 + \lambda^{-a})$ respectively) to find that

$$ \tag{***} f \geq AB - x^a(1+\lambda^a) = x^2 \lambda^{\frac{(2-a)\lambda^a}{1+\lambda^a}} (1+\lambda^a) - x^a(1+\lambda^a)$$

Note that when $a \geq 2$, we have $\ln(\lambda) \cdot \frac{(2-a)\lambda^a}{1+\lambda^a} \geq 0$ for all $\lambda \in (0,1]$, and hence we have

$$ f \geq (x^2 - x^a)(1 + \lambda^a) \geq 0$$

This shows the sharp lower bound when $a \geq 2$ is $c_a = 0$. (The sharpness comes from considering what happens as we take $x$ and $y$ to both approach $0$.)

When $a < 2$

It is again convenient to rewrite in terms of $\beta = \lambda^a$, so we have (using (***) and replacing $\lambda$ by $\beta$)

$$ f \geq x^2 \beta^{\frac{2-a}{a} \frac{\beta}{1+\beta}} (1+\beta) - x^a (1+\beta) $$

Optimize first in $x$ for a fixed $\beta$, we find that the minimum of RHS is attained when $x^{a-2} = \frac2a \beta^{\frac{2-a}{a} \frac{\beta}{1+\beta}}$. Reinserting we see that, for $\beta \in (0,1]$, we have

$$ f \geq \left[ \left(\frac{a}{2}\right)^{\frac{2}{2-a}} - \left(\frac{a}{2}\right)^{\frac{a}{2-a}}\right] \beta^{-\frac{\beta}{1+\beta}}(1+\beta) $$

The function $\beta^{-\frac{\beta}{1+\beta}}(1+\beta)$ attains its maximum at $\beta = 1$, where it takes the value $2$, and hence we have

$$f \geq 2\left[ \left(\frac{a}{2}\right)^{\frac{2}{2-a}} - \left(\frac{a}{2}\right)^{\frac{a}{2-a}}\right] $$

This shows that when $a < 2$, we can bound $c_a \geq 2\left[ \left(\frac{a}{2}\right)^{\frac{2}{2-a}} - \left(\frac{a}{2}\right)^{\frac{a}{2-a}}\right]$. Now, as observed in my previous comment, if you evaluate $f$ along the diagonal $f(x,x,a)$ this becomes a single-variable optimization problem, which can be solved by calculus. And one can check therefore that this lower bound is in fact attained for a well-chosen $x$ (and $y = x$), showing that it is sharp.

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Neither $x$ nor $y$ should be allowed to take the value $0$. So, we have $x$ and $y$ in $(0,1]$. Then, for any real $a>0$, each of the terms $$x,y^{y^a/x^a}, y,x^{x^a/y^a},x^a,y^a$$ takes values in $(0,1]$. So, we have $$c_a\le xy^{y^a/x^a}+yx^{x^a/y^a}-x^a-y^a\le C_a$$ with $c_a=-2$ and $C_a=2$.

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