# Prove that $x^xy^y \geq \dfrac{x^2+y^2}{2}$

## Problem statement.

Let $$x,y$$ be two real positive numbers. Prove that $$x^xy^y\geq\dfrac{x^2+y^2}{2}.$$ (recommend solution with high school knowledge)

## First approach: Bernoulli inequality

1. Case 1: $$x\geq 1, y\geq1$$
I use the Bernoulli inequality that $$(1+x)^r\geq 1+rx,$$ for all $$x>-1$$ and for all $$r\geq1$$. Hence for $$x\geq1$$ we have $$x^x=\left(1+\left(x-1\right)\right)^x\geq 1+x\left(x-1\right)=x^2-x+1.$$ (I am wondering that whether applying is correct or not, because, intuitively, $$x^x$$ is not the same as $$x^r$$. But if I fix for $$x_0\geq0$$, then I can apply Bernoulli inequality for $$x=x_0-1$$ and $$r=x_0$$. Since $$x_0$$ is taken arbitrarily, we obtain the result for all $$x\geq1$$. Please let me know if there is any misunderstood).
So that we have obtain that $$x^xy^x\geq \left(x^2-x+1\right)\left(y^2-y+1\right).$$ We will show that $$\left(x^2-x+1\right)\left(y^2-y+1\right)\geq \dfrac{x^2+y^2}{2}$$ which is equivalent to $$\left(y^2-y+\dfrac{1}{2}\right)x^2-\left(y^2-y+1\right)x+\left(\dfrac{y^2}{2}-y+1\right)\geq0.$$ We have the discriminant $$\Delta = \left(y^2-y+1\right)^2-4\left(y^2-y+\dfrac{1}{2}\right)\left(\dfrac{y^2}{2}-y+1\right)=-\left(y-1\right)^2\leq0.$$ Since $$\left(y^2-y+\frac{1}{2}\right)>0$$, the case for $$x,y\geq 1$$ is proved.

## Second approach: first-order optimality condition

Consider the function $$h(t) = f(t,y):=y^yt^t-\dfrac{t^2}{2}-\dfrac{y^2}{2},$$ for $$t>0$$. Then the first derivative is given by $$f'(t)=y^yt^t\left(\log t+1\right)-t$$ Solving the equation $$f'(t)=0$$ gives $$g(t):=\dfrac{t^{1-t}}{1+\log t}=y^y$$ here we can easily see that $$t=e^{-1}$$ is not a solution. We will show that $$g$$ is decreasing on each well-defined intervals. We have $$g'(t)=-\dfrac{t^{-t}\left(t\left(1+\log t\right)^2-\log t\right)}{\left(1+\log t\right)^2}$$

• If $$t\in(0,1)\setminus\{ e^{-1}\})$$, then $$\log t <0$$, so that $$t\left(1+\log t\right)^2-\log t>0$$, and hence $$g'(t)<0$$.
• If $$t\geq1$$ then $$t\left(1+\log t\right)^2-\log t \geq \left(1+\log t\right)^2-\log t>0$$, and thus $$g'(t)<0$$.
Since $$\lim_{x\to 0^+} g(t)=0$$ and $$lim_{t\to {e^{-1}}^-} g(t) = -\infty$$ and $$lim_{t\to {e^{-1}}^+} g(t)=+\infty$$ and $$\lim_{x\to +\infty} g(t) = 0$$, we can conclude that the equation $$g(t)=y^y$$ has a unique solution, say $$\alpha_y\geq \frac{1}{e}$$. We now obtain that $$f(x,y) \geq h(\alpha_y)=\alpha_y^{\alpha_y}y^y-\dfrac{\alpha_y^2}{2}-\dfrac{y^2}{2}=\dfrac{\alpha_y}{1+\log(\alpha_y)}-\dfrac{\alpha_y^2}{2}-\dfrac{y^2}{2}.$$ Note that $$\alpha_y$$ is dependent of $$y$$, i.e., $$\alpha_y = g^{-1}(y^y)$$. I try to eliminate the exponential term, but it is still difficult for me to handle. Please help me some ideas. Thank you.
• I think you nailed it with your first method. Regarding that, what are the "three remaining cases" you mentioned? Sep 15 at 1:02
• @H.sapiensrex That is $x,y\leq1$ and $(x-1)(y-1)\leq0$ Sep 15 at 6:24
• Great question ..........+1 Sep 19 at 19:17

Here is a proof.

Taking logarithm, it suffices to prove that $$x\ln x + y \ln y \ge \ln\frac{x^2 + y^2}{2}. \tag{1}$$

Using $$\mathrm{e}^u \ge 1 + u$$ for all reals $$u$$, letting $$u = \frac12\ln\frac{x^2 + y^2}{2}$$, we have $$\ln {\frac{x^2 + y^2}{2}} \le 2\sqrt{\frac{x^2 + y^2}{2}} - 2. \tag{2}$$

From (1) and (2), it suffices to prove that $$x\ln x + y \ln y \ge \sqrt{2x^2 + 2y^2} - 2. \tag{3}$$

Note that $$\sqrt{2x^2 + 2y^2} - (x + y) = \frac{(x - y)^2}{\sqrt{2x^2 + 2y^2} + x + y} \le \frac{(x - y)^2}{x + y + x + y}. \tag{4}$$

From (3) and (4), it suffices to prove that $$x\ln x + y \ln y \ge x + y + \frac{(x - y)^2}{2x + 2y} - 2. \tag{5}$$

To proceed, we need the following result.

Fact 1. For all $$u > 0$$, $$u\ln u \ge \frac{(5u+1)(u-1)}{2u+4}.$$ (Note: The RHS is the $$(2, 1)$$ Pade approximation at $$u = 1$$ of $$u\ln u$$.)
(Proof: Let $$h(u) := \ln u - \frac{1}{u}\cdot\frac{(5u+1)(u-1)}{2u+4}$$. We have $$h'(u) = \frac{(u-1)^3}{u^2(u+2)^2}$$. Thus, $$h(u)$$ is strictly decreasing on $$(0, 1)$$, and strictly increasing on $$(1, \infty)$$. Note also that $$h(1) = 0$$. Thus, we have $$h(u)\ge 0$$ for all $$u > 0$$. )

Using Fact 1, it suffices to prove that $$\frac{(5x+1)(x-1)}{2x+4} + \frac{(5y+1)(y-1)}{2y+4} \ge x + y + \frac{(x - y)^2}{2x + 2y} - 2$$ or \begin{align*} &2{x}^{3}y + 8{x}^{2}{y}^{2} + 2x{y}^{3} + 4{x}^{3} - 4{x}^{2}y -4x{ y}^{2}+4{y}^{3}\\ &\qquad -13{x}^{2}-10xy-13{y}^{2}+12x+12y\\ \ge{}& 0. \tag{6} \end{align*}

Let $$p = x + y, q = xy$$. Then $$p^2 \ge 4q$$. (6) is written as $$4q^2 + (2p^2 - 16p + 16)q + 4p^3 - 13p^2 + 12p \ge 0.$$ The rest is smooth.

Actually, (6) can be written as \begin{align*} &{\frac { \left( 4\,{x}^{2}+8\,xy+4\,{y}^{2}-13\,x-13\,y+12 \right) \left( x-y \right) ^{4}}{ \left( x+y \right) ^{3}}}\\ &\qquad +{\frac { 2\left( x+y+12 \right) \left( x+y-2 \right) ^{2} \left( x-y \right) ^ {2}xy}{ \left( x+y \right) ^{3}}}\\ &\qquad +{\frac { 12\left( x+y+4 \right) \left( x+y-2 \right) ^{2}{x}^{2}{y}^{2}}{ \left( x+y \right) ^{3}}} \ge 0 \end{align*} which is clearly true.

We are done.

WLOG, assume that $$0 < x \le y$$.

Case 1: $$x \ge 1$$

By Bernoulli inequality, we have $$x^x \ge 1 + (x - 1)x = x^2 - x + 1 \ge \frac{x^2 + 1}{2}.$$

We have $$x^xy^y - \frac{x^2 + y^2}{2} \ge \frac{x^2 + 1}{2}\frac{y^2 + 1}{2} - \frac{x^2 + y^2}{2} = \frac{(x^2 - 1)(y^2 - 1)}{4} \ge 0.$$

Case 2: $$y \le 1$$

By Bernoulli inequality, we have $$y^y = \frac{y}{y^{1 - y}} \ge \frac{y}{1 + (y - 1)(1 - y)} = \frac{1}{2 - y} \ge \frac{y^2 + 1}{2}$$ where we use $$\frac{1}{2 - y} - \frac{y^2 + 1}{2} = \frac{y(y - 1)^2}{2(2 - y)} \ge 0$$.

We have $$x^xy^y - \frac{x^2 + y^2}{2} \ge \frac{x^2 + 1}{2}\frac{y^2 + 1}{2} - \frac{x^2 + y^2}{2} = \frac{(x^2 - 1)(y^2 - 1)}{4} \ge 0.$$

• Thank you for your help. It seems we didn’t use the assumption $x\leq y$. Two your partial cases gives a proof that $x^x\geq \frac{x^2+1}{2}$ for $x>0$. Sep 15 at 6:27
• @Chivul Don't accept it since it is partial answer. You can upvote it if you think it is helpful. We use $x \le y$ in Bernoulli inequality. Indeed, fact 1 is $1 \le x \le y$, fact 2 is $0 < x \le y \le 1$. For example, in case 1, we need $y \ge 1$ to use $y^y \ge 1 + (y - 1)y \ge \frac{y^2 + 1}{2}$ similarly. Sep 15 at 6:30
• ah, I see. I forgot that we also apply the Bernoulli inequality for the other variable. Sep 15 at 6:33
• @Chivul Yes, it is. The remaining case is that $0 < x < 1 < y$. Sep 15 at 6:34
• but we now can prove that $x^x\geq (x^2+1)/2$ for all $x>0$. This is a result as a lemma, not relate to $x,y$ this case. So we use it no matter the case. Sep 15 at 6:37

Another Proof. By taking logarithm to both sides, we may instead prove

$$x \log x + y \log y \quad \color{red}{\geq} \quad \log\left(\frac{x^2+y^2}{2}\right). \tag{1}$$

To this end, we substitute $$(x, y) = (\lambda p, \lambda q)$$, where $$\lambda > 0$$ and $$p, q \in (0, 1)$$ with $$p + q = 1$$. Then a bit of algebra tells that $$\text{(1)}$$ is equivalent to

$$\bbox[color:navy;padding:5px;border:1px dotted navy;]{(\lambda - 2)\log(\lambda/2) + D \lambda - E \quad \color{red}{\geq} \quad 0,} \tag{2}$$

where $$D$$ and $$E$$ are defined by

\begin{align*} D &= \log 2 + p\log p + q\log q, \\ E &= \log 2 + \log(p^2 + q^2). \end{align*}

(These quantities can be interpreted as a Rényi divergence of order $$1$$ and $$2$$, respectively, although this observation is only tangentially used to discover inequalities between $$D$$ and $$E$$ here.)

In order to facilitate this parametrization, we investigate the properties of $$D$$ and $$E$$.

Lemma 1. Let $$\delta = p - q \in (-1, 1)$$. Then $$D \geq \delta^2/2$$ and $$E = \log(1+\delta^2)$$.

Lemma 2. We have $$0 \leq D \leq E \leq 2D$$.

Lemma 3. We have $$(\lambda - 2)\log(\lambda/2) \geq \frac{1}{2}(\lambda - 2)^2$$ for $$\lambda \in (0, 2]$$.

The proof of these lemmas is mostly straightforward and will be illustrated at the end of this answer. Now let's see how this helps prove OP's claim.

$$\boxed{\text{Case 1}}$$ Suppose $$\lambda \geq 2$$. Then

$$\text{[LHS of (2)]} \geq D\lambda - E \geq 2D - E \geq 0,$$

where we invoked Lemma 2 in the last step.

$$\boxed{\text{Case 2}}$$ Suppose $$\lambda \in (0, 2]$$. Then applying Lemma 2 and completing the square,

\begin{align*} \text{[LHS of (2)]} &\geq \frac{1}{2}(\lambda - 2)^2 + D\lambda - E \tag{Lemma 3} \\ &= \frac{1}{2}(\lambda - 2 + D)^2 + 2D - \frac{D^2}{2} - E \\ &\geq 2D - \frac{D^2}{2} - E. \end{align*}

Now by the lemmas and the inequality $$e^x - 1 \geq x + \frac{x^2}{2}$$, $$x \geq 0$$, together,

\begin{align*} 2D - \frac{D^2}{2} - E &\geq 2D - \frac{E^2}{2} - E \tag{Lemma 2} \\ &\geq 2D - (e^E - 1) \\ &\geq \delta^2 - (e^{\log(1+\delta^2)} - 1) \tag{Lemma 1} \\ &= 0. \end{align*}

Conclusion. Therefore $$\text{(2)}$$ holds unconditionally and we are done. $$\blacksquare$$

Proof of Lemma 1. Let us regard $$D$$ and $$E$$ as functions of $$\delta$$. Plugging $$p = \frac{1+\delta}{2}$$ and $$q = \frac{1-\delta}{2}$$ into the definition of $$D$$ and $$E$$, we get

$$D = \frac{(1+\delta)\log(1+\delta) + (1-\delta)\log(1-\delta)}{2}, \qquad E = \log(1+\delta^2).$$

From this, we can easily check that $$D(0) = D'(0) = 0$$ and $$D''(\delta) = \frac{1}{1-\delta^2} \geq 1$$, which proves the desired assertion.

Proof of Lemma 2. Using Lemma 1 and the inequality $$\log(1+x) \leq x$$ for $$x > 0$$, we get

$$0 \leq \delta^2/2 \leq D \qquad\text{and}\qquad E \leq \delta^2 \leq 2D.$$

On the other hand, the map $$x \mapsto \log(2x)$$ is concave, hence by Jensen's inequality,

$$D = p\log(2p) + q\log(2q) \leq \log(2p^2 + 2q^2) = E.$$

Proof of Lemma 3. Let $$f(\lambda) = (\lambda - 2)\log(\lambda/2)$$. This function satisfies $$f(2) = f'(2) = 0$$ and $$f''(\lambda) = \frac{2}{\lambda^2} + \frac{1}{\lambda} \geq 1$$ for $$\lambda \in (0, 2]$$, proving the desired claim.

Addendum. Just for fun, I also included an animated figure demonstrating a variant of $$\text{(2)}$$,

$$(\lambda - 2)\log(\lambda/2) \geq - D \lambda + E$$

as functions of $$\lambda$$, for various values of $$p$$: • It is interesting. (+1) Sep 19 at 22:43
• @RiverLi, Thank you! :) Sep 20 at 5:26

Supplement to @Bob Dobbs's nice idea.

Let $$f(x, y) := 2 - \frac{x^2 + y^2}{x^x y^y}$$.

Fact 1. If $$x, y > 0$$ with $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$$, then $$x = y = 1$$. (The proof is given at the end.)

Fact 1 tells us that $$x = y = 1$$ is the only stationary point.

$$\phantom{2}$$

Proof of Fact 1.

We have \begin{align*} \frac{\partial f}{\partial x} &= \frac{(x^2 + y^2)(\ln x + 1) - 2x}{x^xy^y} = 0, \tag{1}\\ \frac{\partial f}{\partial y} &= \frac{(x^2 + y^2)(\ln y + 1) - 2y}{x^xy^y} = 0. \tag{2} \end{align*}

From (1) and (2), we have \begin{align*} \frac{\ln x + 1}{x} - \frac{\ln y + 1}{y} &= 0, \tag{3}\\ (x^2 + y^2)(\ln x - \ln y) - 2(x - y) &= 0. \tag{4} \end{align*}

Let us prove that $$x = y = 1$$ is the only solution to the system of equations (3) and (4).

Assume, for the sake of contradiction, that $$x > y > 0$$ is also a solution.

Let $$r = \frac{x}{y} > 1$$. Plugging $$x = ry$$ into (3), we have $$\frac{\ln r - (r - 1) - (r - 1)\ln y}{ry} = 0$$ which results in $$y = \mathrm{e}^{-1}r^{\frac{1}{r - 1}}.$$ Then $$x = ry = \mathrm{e}^{-1}r^{\frac{r}{r - 1}}$$.

Thus, we have \begin{align*} &(x^2 + y^2)(\ln x - \ln y) - 2(x - y)\\ ={}& (r^2 + 1)y^2\ln r - 2(r - 1)y\\[6pt] >{}& \frac{(r + 1)^2}{2}\cdot y^2\cdot \frac{2(r - 1)}{r + 1} - 2(r - 1)y\\[6pt] ={}&\mathrm{e}^{-1}(r + 1)(r - 1)y\left( r^{\frac{1}{r - 1}}- \frac{2\mathrm{e}}{r + 1}\right)\\ >{}& 0 \tag{5} \end{align*} where we use $$r^2 + 1 > \frac{(r + 1)^2}{2}$$, and $$\ln r > \frac{2(r - 1)}{r + 1}$$ for all $$r > 1$$ (easy to prove), and $$\ln (r^{\frac{1}{r - 1}}) - \ln\frac{2\mathrm{e}}{r + 1} = \frac{\ln r}{r - 1} - \ln\frac{2\mathrm{e}}{r + 1} > \frac{\frac{2(r - 1)}{r + 1}}{r - 1} - 1 - \ln\frac{2}{r + 1} > 0$$ where we use $$\ln v < v - 1$$ for all $$v\in (0, 1)$$ (easy to prove).

We are done.

We want to show that $$f(x,y)=x^{2-x}y^{-y}+x^{-x}y^{2-y}\leq 2$$ for $$x,y\in\Bbb R^{+}$$. The critical points are given by $$f_x=f_y=0$$ and hence $$\frac{x}{1+\ln x}=\frac{y}{1+\ln y}\tag1$$ $$\frac{\ln x-\ln y}{x-y}=\frac{2}{x^2+y^2}\tag2$$ By $$(1)$$, we have two cases.

1. If $$y=x$$, it is not difficult to show that the only solution of the system is $$x=y=1$$.
2. If they are different, without loss of generality $$x>1>y$$ and by using $$(1)$$ and $$(2)$$, we have $$\frac{\ln x-\ln y}{x-y}=\frac {1+\ln x}{x}=\frac{2}{x^2+y^2}<\frac{2}{x^2}$$ which is not true for $$x\geq1.46$$.

For $$1: The tangent line of the curves $$\frac{\ln x +1}{x}=\frac2{x^2+y^2}$$ and $$\frac{\ln y +1}{y}=\frac2{x^2+y^2}$$ as shown in WolframAlpha at the point $$(1,1)$$ is $$y=2-x.$$ Now, it is enough to show that the tangent line does not intersect the curve $$\frac{\ln x +1}{x}=\frac2{x^2+y^2}$$. So, it is enough to show that the curve $$y=\ln x+1-\frac{x}{x^2+(x-2)^2}$$ has no zeros for $$x>1.$$ From $$y'>0$$, we get $$(x-1)^4+x^3+2x^2-6x+3>0$$ for $$x>1.$$ We are done.

• I think, $x$ and $y$ can not be greater (less) than $1$ at the same time due to $(1)$. @RiverLi Sep 19 at 0:48
• I think you need to describe it more clearly. Sep 19 at 0:56
• I found that $\frac{\ln x-\ln y}{x-y}=\frac{2}{x^2+y^2}$ has a real root satisfying $0 < y < 1 < x$. For example, $y = 9/10$, and $x \approx 1.078465394$ (perhaps the exact value of $x$ does not admit a closed form). You may check. So if you only use $x$ and $y$ can not be greater (less) than 1 at the same time, it is not enough. Sep 19 at 1:01
• Maybe, the condition is $1/x$ must be between them. Sep 19 at 1:18
• +1 for this nice answer Sep 19 at 19:17

Using $$e^x\ge x+1$$ for a real $$x$$

We have :

$$f(x)=x^x\ge x((x-1)\ln(x)+1)\ge h(x)=x(2(x-1)^2/(x+1)+1) ,\forall x>1$$

Remains to show $$x\ge 1,y\in(0,1)$$:

$$h(x)f(y)\geq r(x,y)=1/2(x^2+y^2)$$

Where we can use derivative.

Proof of the hardest case $$0 :

We have to show $$x\to xa,y\to a$$ :

$$xa\left(\frac{2\left(xa-1\right)^{2}}{xa+1}+1\right)a^{a}-\frac{\left(1+x^{2}\right)a^{2}}{2}\geq 0$$

Or :

$$x\left(\frac{2\left(xa-1\right)^{2}}{xa+1}+1\right)a^{1+a}-\frac{\left(1+x^{2}\right)a^{2}}{2}\geq 0$$

Or using binomial inequality :

$$b(x,a)=x\left(\frac{2\left(xa-1\right)^{2}}{xa+1}+1\right)\left(1+\left(a+1\right)\left(a-1\right)+\frac{1}{2}\left(a-1\right)^{2}a\left(a+1\right)+\frac{1}{6}\left(a-1\right)^{4}\left(a+1\right)\left(a\right)\right)-\frac{\left(1+x^{2}\right)a^{2}}{2}\geq0$$

Then using a computer :

$$b\left(\left(x+y+1\right)^{2},\frac{1}{y+1}\right)\geq 0$$ all coefficient positive

This ugly but the advantage we can set all coefficient equal to one and get a nice inequality .

Come back for the trivial case .

Trivial case :

New bound for Am-Gm of 2 variables here I show for positive $$x,y$$ the inequality :

$$x\ln x +y\ln y\geq (x+y)\ln(x+y-\sqrt{xy})$$

Without derivative. Now we apply Bernoulli's to one of the side Remains to show

Remaing proof for $$x+y\geq 1$$:

Let $$x,y>0$$ then we have :

$$1+\left(x+y\right)\left(x+y-\sqrt{xy}-1\right)\geq\frac{x^{2}+y^{2}}{2}$$

Setting : $$\alpha\geq 2,x=\alpha/2u,y=\alpha/(2x)$$

Then we need to show :

$$1+\frac{\alpha}{2}\left(u+\frac{1}{u}\right)\left(\frac{\alpha}{2}\left(u+\frac{1}{u}\right)-\frac{\alpha}{2}-1\right)\geq\frac{\frac{\alpha^{2}}{4}\left(u^{2}+\frac{1}{u^{2}}\right)}{2}=\frac{\left(\frac{\alpha^{2}}{4}\left(u+\frac{1}{u}\right)^{2}-\frac{\alpha^{2}}{2}\right)}{2}$$

Setting $$\frac{\alpha}{2}\left(u+\frac{1}{u}\right)=a,b=\frac{\alpha}{2}=b$$:

$$1+a\left(a-b-1\right)\ge \frac{a^{2}}{2}-b^{2}$$

Or :

$$1+\frac{a^{2}}{2}-ab-a+b^{2}\geq 0$$

Or :

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

Now starting from :

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

Or :

$$1+\left(a-b\right)^{2}+1+b^{2}-2a\geq 0$$

Or using am-gm :

$$1+\left(a-b\right)^{2}+2b-2a\geq 0$$

Or :

$$(a-b-1)^2\geq 0$$

For the case $$x+y\le 1$$

using Jensen's inequality because $$x\ln x$$ is convex over the positive real we need to show :

$$\left(\frac{\left(x+y\right)}{2}\right)^{\left(x+y\right)}-\frac{\left(x^{2}+y^{2}\right)}{2}\geq 0$$

Or :

$$\left(\frac{\left(x+y\right)}{2}\right)^{\left(x+y\right)}-\frac{x}{2}-\frac{y}{2}>0$$

Or :

$$u=x+y,(u/2)^u-u/2\geq 0$$

Which is trivial when $$u\in(0,1]$$

We are done .

Conjecture :

Let $$x,y>0,n>2,x\neq y$$ then it seems we have :

$$\left(\left(1+\frac{e^{2n}\left(x-y\right)^{2}}{\left(x+y\right)^{2}}\right)x^{x}y^{y}\right)-\frac{\left(x^{n}+y^{n}\right)^{\frac{2}{n}}}{2^{\frac{2}{n}}}>0$$

• So this is a proof only for the case $xy \ge 1$? – Unless I am mistaken, the last inequality does not hold for $x=100, y=1/100$. Sep 19 at 7:49
• @MartinR corrected and true I can add details Sep 19 at 10:04
• $0 < y < 1 \le x$ seems to be the difficult case. It would be nice to see a complete answer, not a vague hint like “use derivative.” Sep 19 at 11:24
• @MartinR can you criticize ? Sep 21 at 8:58

The trivial case $$x,y\in[1,\infty)\operatorname{or} x,y\in(0,1]$$:

We have the inequality for $$x>0$$ :

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

Now we need to show :

$$h(x)h(y)-\left(\frac{1}{2x^2}+\frac{1}{2y^2}\right)^{-1}\leq 0$$

Or :

$$-2 y^2 (y - 1) (y + 1) x^2 (x - 1) (x + 1) / ((y^2 + 1) (x^2 + 1) (x^2 + y^2))\leq 0$$

For the hard case we have $$0:

Let $$x\in(0,1]$$ :

$$x^{\frac{1}{x}}\leq g(x)=4x^3/(x^2+1)$$

Then define the difference :

$$a(x,y)=g(x)h(y)-\left(\frac{1}{2x^2}+\frac{1}{2y^2}\right)^{-1}$$

Then :

$$a(x,y)=-2 x^2 (x - 1) y^2 (y^2 x^3 + y^2 x^2 + 3y^2 x - y^2 + x^3 - 3x^2 - x - 1) / ((x^2 + 1)^2 (y^2 + 1) (y^2 + x^2))\le 0$$

Provided :

$$y^2 x^3 + y^2 x^2 + 3y^2 x - y^2 + x^3 - 3x^2 - x - 1\leq 0$$

Case $$xy\leq 1$$ we have :

$$\frac{16x^3y^3}{(x^2+1)^2(y^2+1)^2}\ge f(x)f(y)$$

The difference with the upper bound is equal to :

$$-2 y^2 x^2 (x^4 y^4 + 2x^4 y^2 + x^4 - 8x^3 y + 2x^2 y^4 + 4x^2 y^2 + 2x^2 - 8x y^3 + y^4 + 2y^2 + 1) / ((y^2 + 1)^2 (x^2 + 1)^2 (x^2 + y^2))\le 0$$