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

Question

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)

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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.
   
2. I meet the stuck with three remaining cases. Please help me if you would like to follow this approach. Thank you.

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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.

Answer 1

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.

Answer 2

Partial answer.

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

Answer 3

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 $1<x<1.46$: 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.

Answer 4

Complete answer

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 $0<y\leq 1\leq x$ :

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 :

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

Answer 5

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. $\blacksquare$

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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.

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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 $p$:

Answer 6

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.

$\phantom{2}$

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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*}

If $x = y$, from (1), we have $2x^2(\ln x + 1) - 2x = 0$ or $\ln x + 1 - \frac{1}{x} = 0$ which results in $x = 1$.
(Note: Let $f(x) := \ln x + 1 - \frac{1}{x}$. Then $f'(x) = \frac{1}{x} + \frac{1}{x^2} > 0$ and $f(1) = 0$.)

So it suffices to prove that there is no $x, y > 0$ with $x \ne y$ satisfying the system of equations (1) and (2).

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*}

It suffices to prove that there is no $x, y > 0$ with $x \ne y$ satisfying the system of equations (3) and (4).

Assume, for the sake of contradiction, that $x > y > 0$ satisfies the system of equations (3) and (4).

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).

However, (5) contradicts (4).

We are done.

Answer 7

Hint as exercise :

Case $x+y\geq 2,x\neq y$ :

1. Show that $$x^{\frac{x}{x+y}}y^{\frac{y}{y+x}}\sqrt{\frac{\left(x^{2}+y^{2}\right)}{2}}>\frac{\left(x^{2}+y^{2}\right)}{2}$$

2. Suppose $x^{\frac{x}{x+y}}y^{\frac{y}{y+x}}\sqrt{\frac{\left(x^{2}+y^{2}\right)}{2}}>\frac{\left(x^{2}+y^{2}\right)}{2}>x^{x}y^{y}$ use AM-GM and show that :

$$x^{\frac{x}{x+y}}y^{\frac{y}{y+x}}<x^{\frac{x}{2}}y^{\frac{y}{2}}\tag{I}$$

2. b Using Bernoulli's inequality show for $x> 0,y\in(-1,0)$ :

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

3. Conclude for this case

Case $x+y\leq 2,x\neq y$ :

The goal of this part is to show the inequality for $x\in[0.25,1.75]$ :

$$x^{x}y^{y}>\frac{\left(x^{2}+y^{2}\right)}{2}\left(\frac{5}{4}\left(1-x^{\frac{x}{2}}y^{\frac{y}{2}}\right)^{2}+1\right)$$

We introduce the function :

$$f\left(x\right)=1-x^{-x}y^{-y}\frac{\left(x^{2}+y^{2}\right)}{2}\left(\frac{5}{4}\left(1-x^{\frac{x}{2}}y^{\frac{y}{2}}\right)^{2}+1\right)$$

Show that :

$$f'(x)\leq 0,x\in[0.25,1.75]$$

3)Show the last inequality which is a one unknow for $x\in[0.25,1.75]$ .

4)Conclude

Answer 8

Sketch of proof:

We need to show that

$$(1+x)^{1+x} \cdot (1+y)^{1+y} \ge \frac{(1+x)^2 + (1+y)^2}{2}$$

for $x$, $y\ge -1$. We can do it in two steps:

1. Prove that for $x\ge -1$ we have the inequality:

$$(1+x)^{1+x} \ge 1 + x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \frac{x^5}{12} \ ( \ge 0)$$

2. Prove that for $x$, $y\ge -1$ we have

$$(1 + x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \frac{x^5}{12}) \cdot (1 + y + y^2 + \frac{y^3}{2} + \frac{y^4}{3} + \frac{y^5}{12}) \ge \\ \ge \frac{(1+x)^2 + (1+y)^2}{2}$$

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$\bf{Added:}$ Proving the inequality 1. : Note he Taylor series expansion at $x=0$
$$(1+x)^{1+x} = \exp( (1+x) \cdot \log(1+x)) = 1+x+x^2 + \frac{x^3}{2}+\frac{x^4}{3} + \frac{x^5}{12} +\frac{3 x^6}{40}- \cdots$$

So at least around $0$ this is a fairly good inequality. To show it for all $x\ge -1$ consider the equivalent inequality

$$f(x) \colon = (1+x) \log(1+x) - \log(1+x+x^2 + \frac{x^3}{2}+\frac{x^4}{3} + \frac{x^5}{12}) \ge 0$$

It is enough to show that $f$ is convex on $[-1, \infty)$, since that implies $0$ is a point of minimum, with $f'(0) = 0$. Now we calculate:

$$f''(x) = \frac{x^4 (324 + 468 x + 364 x^2 + 192 x^3 + 65 x^4 + 13 x^5 + x^6)}{(1 + x) (12 + 12 x + 12 x^2 + 6 x^3 + 4 x^4 + x^5)^2}$$

and the expression on RHS is $\ge 0$ for $x\ge -1$.