show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$

Question

let $a,b>0.$ Show that $$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\tag{1}$$

I known How to prove $a^b+b^a>1$,where $a,b>0.$ See $x^y+y^x>1$ for all $(x, y)\in \mathbb{R_+^2}$

to prove $（1）$, I want use AM-GM inequality $$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\left(\dfrac{a^b}{b}\cdot\dfrac{b^a}{a}\right)^{1/4}=2\left(a^{b-1}b^{a-1}\right)^{1/4}$$

But $a^{b-1}b^{a-1}$ is not always $>1$

Answer 1

Remarks (2022/04/01): @Erik Satie's proof is simpler than mine.

Sketch of @Erik Satie's proof:

Fact 3: Let $0 < y \le 1 \le x$. Then $$x^{y^2} \ge 1 + \left(x \cdot \frac{1}{1 - (x - 1)(y - 1)} - 1\right) y.$$

Fact 4: Let $0 < y \le 1 \le x$. Then $$y^{x^2} \ge 1 + \left(y \cdot \frac{1}{1 - (x - 1)(y - 1)} - 1\right) x.$$

By Facts 3-4, it suffices to prove that $$\frac{1}{x}\left[1 + \left(\frac{x}{1 - (x - 1)(y - 1)} - 1\right) y\right] + \frac{1}{y}\left[1 + \left(\frac{y}{1 - (x - 1)(y - 1)} - 1\right) x\right] \ge 2$$ or $$\frac{(2xy - x - y)^2}{(x + y - xy)xy} \ge 0$$ which is true.

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Update: I found a simpler proof.

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WLOG, assume that $b \le a$.

If $a, b > 1$ or $a, b < 1$, then $a^{b - 1}\ge 1$ and $b^{a - 1}\ge 1$ and thus $$\sqrt{\frac{a^b}{b}} + \sqrt{\frac{b^a}{a}}\ge 2\sqrt[4]{a^{b-1}b^{a-1}} \ge 2.$$

It remains to prove the case when $0 < b \le 1 \le a$.

Let $a = x^2, b = y^2$. It suffices to prove that, for all $0 < y \le 1 \le x $, $$\frac{x^{y^2}}{y} + \frac{y^{x^2}}{x} \ge 2.$$

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Fact 1: If $x \ge 1$ and $0 < y \le 1$, then $$x^{y^2} \ge \frac{1 + x + (x - 1)y^2}{1 + x - (x - 1)y^2}.$$ (Proof: Let $f(x) = y^2\ln x - \ln \frac{1 + x + (x - 1)y^2}{1 + x - (x - 1)y^2} $. We have $f'(x) = \frac{(1 - y^4)y^2(x - 1)^2}{x[(1 + x)^2 - (x - 1)^2y^4]}\ge 0$. Also, $f(1) = 0$. Thus, $f(x) \ge 0$ for all $x\ge 1$.)

Fact 2: If $x \ge 1$ and $0 < y \le 1$, then $$y^{x^2} \ge \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2}.$$ (The proof is given at the end.)

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Now, using Facts 1-2, it suffices to prove that $$\frac{1}{y}\cdot \frac{1 + x + (x - 1)y^2}{1 + x - (x - 1)y^2} + \frac{1}{x}\cdot \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2} \ge 2.$$

Let $x = 1 + s$ for $s \ge 0$. After clearing the denominators, it suffices to prove that $$q_4 s^4 + q_3 s^3 + q_2 s^2 + q_1s + q_0 \ge 0 \tag{1}$$ where \begin{align*} q_4 &= (1 - y)(2y^3 + y^2 - 2y + 1), \\ q_3 &= (1 - y)(7y^3 + 3y^2 - 11y + 5), \\ q_2 &= - 6y^4 + 8y^3 + 24y^2 - 32y + 10, \\ q_1 &= -2y^4 + 4y^3 + 16y^2 - 28y + 10, \\ q_0 &= 4(1 - y)^2. \end{align*} It is easy to prove that $q_4, q_3, q_2, q_0 \ge 0$. Also, we have \begin{align*} 4q_2q_0 - q_1^2 = 4(y^3 + y^2 + 7y + 15)(1 - y)^5 \ge 0. \end{align*} Thus, (1) is true.

We are done.

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Proof of Fact 2: We only need to prove the case when $\frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2} > 0$, i.e. $y > \frac{x^2 - 1}{x^2 + 1}$. In other words, we only need to prove the case when $\frac{x^2 - 1}{x^2 + 1} < y \le 1$. Let $$g(y) = x^2\ln y - \ln \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2}.$$ We have $$g'(y) = - \frac{x^2(x^4 - 1)(1 - y)^2}{y[1 + y + (y - 1)x^2]^2}\cdot \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2} \le 0.$$ Also, $g(1) = 0$. Thus, $g(y) \ge 0$ for all $y \in (0, 1]$.

We are done.

Answer 2

My try (please point out the errors if any!):

Multiplying both sides by $\sqrt{ab}$, we get that $$\sqrt{a^{b+1}}+\sqrt{b^{a+1}}\geq 2\sqrt{ab}\tag{1}$$ $$\frac{\sqrt{a^{b+1}}+\sqrt{b^{a+1}}}{2}\geq \sqrt{ab}$$ Using AM-GM inequality,

$$\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\sqrt{ab}}$$

For $0\leq a\leq b\leq 1$, we have that $\sqrt{\sqrt{ab}}\geq\sqrt{ab}$, $\sqrt{a^{b+1}}\geq \sqrt{a}$ and $\sqrt{b^{a+1}}\geq \sqrt{b}$. Therefore, $$\frac{\sqrt{a^{b+1}}+\sqrt{b^{a+1}}}{2}\geq\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\sqrt{ab}}\geq \sqrt{ab}$$

Thus, the original inequality holds.

For $1\leq a\leq b$, we can transform $(1)$ into $$a^{b+1}+b^{a+1}+2{a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\ge 4ab$$

As ${a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\geq ab$, it follows that $4ab-2{a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\leq 2$. As $a^{b+1}+b^{a+1}\geq 2$, the original inequality holds in this case too.

QED

Answer 3

Alternative sketch of proof for $0<x\leq 1\leq a$

Using derivative we have the inequalities :

$$\left(1+\left(a\cdot\frac{1}{1-\left(a-1\right)\left(x-1\right)}-1\right)x\right)\leq a^{x^2}$$

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

Then we need to show :

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

Wich is easy !

Answer 4

My second proof:

We use isolated fudging.

It suffices to prove that, for all $a, b > 0$, $$\sqrt{\frac{a^b}{b}} \ge \frac{3\sqrt{a} - \sqrt{b}}{\sqrt{a } + \sqrt{b}}. \tag{1}$$ (Summing cyclically on (1), the desired result follows. The proof of (1) is given at the end.)

We are done.

$\phantom{2}$

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Proof of (1):

Letting $x = \sqrt{a/b}$, the desired inequality is written as $$b^{(b-1)/2}x^b \ge \frac{3x-1}{x+1}.$$

We only need to prove the case that $x > 1/3$.

Taking logarithm, it suffices to prove that, for all $b > 0$ and $x > 1/3$, $$f(b, x) := \frac{b-1}{2}\ln b + b\ln x - \ln \frac{3x-1}{x+1} \ge 0. $$

It is not difficult to prove that $f(b, x) \ge 0$ on the boundary of $b> 0, x > 1/3$.

It remains to prove that $f(b, x) \ge 0$ for all stationary points in the interior of the region.

We have \begin{align*} \frac{\partial f}{\partial b} &= \frac12\ln b + \frac{b-1}{2b} + \ln x, \\[6pt] \frac{\partial f}{\partial x} &= \frac{(x+1)(3x-1)b - 4x}{(3x-1)(x+1)x}. \end{align*}

We claim that $\frac{\partial f}{\partial b} = \frac{\partial f}{\partial x} = 0$ has exactly one solution $(b, x) = (1, 1)$ on $b > 0, x > 1/3$. Indeed, from $\frac{\partial f}{\partial x} = 0$, we have $b = \frac{4x}{(x+1)(3x-1)}$. Inserting $b = \frac{4x}{(x+1)(3x-1)}$ into $\frac{\partial f}{\partial b} = 0$, we have $$h(x) := \ln\frac{4x^3}{(3x-1)(x+1)} - \frac{(3x+1)(x-1)}{4x} = 0.$$ We have $$h'(x) = -\frac{(9x^2+12x-1)(x-1)^2}{4x^2(3x-1)(x+1)}.$$ Thus, we have $h'(x) < 0$ on $(1/3, 1) \cup (1, \infty)$. Also, we have $h(1) = 0$. Thus, $h(x) = 0$ has exactly one solution $x = 1$. Then $b = 1$. The claim is proved.

Note that $f(1, 1) = 0$. Thus, we have $f(b, x) \ge 0$ for all stationary points in the interior of the region.

We are done.

Answer 5

Using again Generalized Young inequality we have :

$$\left(\frac{1}{a^{b}}+\frac{1}{x^{b}}\right)\left(a^{\sqrt{2^{-1}}x^{\left(1-b\right)}}\cdot x^{\left(\sqrt{2^{-1}}\left(a\right)^{\left(1-b\right)}\right)}\cdot a^{\frac{\left(b-0.5\right)\sqrt{2}}{a^{b}}}\cdot x^{\frac{\left(b-0.5\right)\sqrt{2}}{x^{b}}}\right)^{\frac{1}{\sqrt{2}a^{-b}+\frac{\sqrt{2}}{x^{b}}}}\leq \left(\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}\right)$$

where $b\to1$ and $a,x>0$

Final conjecture :

Let $a,x>0$ then it seems we have :

$$\left(\frac{1}{a^{b}}+\frac{1}{x^{b}}\right)\left(a^{\sqrt{2^{-1}}x^{\left(1-b\right)}}\cdot x^{\left(\sqrt{2^{-1}}\left(a\right)^{\left(1-b\right)}\right)}\cdot a^{\frac{\left(b-0.5\right)\sqrt{2}}{a^{b}}}\cdot x^{\frac{\left(b-0.5\right)\sqrt{2}}{x^{b}}}\right)^{\frac{1}{\sqrt{2}a^{-b}+\frac{\sqrt{2}}{x^{b}}}}\geq 2$$

Where $b\to 1$

I think we can settle $b=1$ it works also .

Using a bit of algebra (introducing log and invert the variables)

We need to show :

$x,a > 0$, we need to prove that $$f\left(x\right)=(x+a)\ln\frac{a+x}{2}-\frac{\left(a+1\right)\ln a+\left(x+1\right)\ln x}{2} \ge 0.$$

A proof of this fact can be found here :

Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$

Answer 6

COMMENT.- I have trouble writing in English so this schematic presentation.

► $f(x,y)=\sqrt{\dfrac{x^y}{y}}$

► One has to prove for all point $(a,b)$ of the first quadrant $$f(a,b)+f(b,a)\ge 2$$ ► For all positive value $k$ the curves $f(x,y)=k$ and $f(y,x)=k$ are symmetric respect to the line $y=x$.

► One can use the black curve above to reduce the problem to prove for the complement of the region defined by $$f(x,y)\ge 2$$ because for any point in the shadow region the proposed inequality is trivially verified (one term of the sum, $f(x,y)$, is already greater than $2$). Note that a neighborhood of $0$ is discarded which is “natural” because a denominator very small is involved.

► Now we can reduce the white region with the symmetric function $f(y,x)=\sqrt{\dfrac{y^x}{x}}=2$ (note for all point $(b,a)$ “above” this curve its symmetric point $(a,b)$ satisfies $f(a,b)\gt2$ so $f(a,b)+f(b,a)\ge 2$).

► So it remains to prove the inequality for the points inside the white region.

HINT.- Let the line $L: y = -x + a$ where $a$ runs through the interval $[0.36,5.02]$ (these numbers correspond to the values of $a$ for the points of intersection of the two considered symmetric curves). We are going to prove that each point $(a,b)$ of the segment $\overline{PQ}$, where $P$ and $Q$ are the points of intersection of the line with said curves, satisfies the inequality.

For this, let $F$ be the function $$F(x)=\sqrt{\frac{x^{a-x}}{a-x}}+\sqrt{\frac{(a-x)^x}{x}}$$ and study its variation. For example, for $a=4$, if the point $P=(x_0,y_0)$ then $x_0\approx 0.426$ and $F(0.426)\approx2.1247\gt2$ (ideally obviously must be equal to $2$). At the point where $x=y$, when $x=2$ (because of $x=-x+4$) one has $F(2)\approx2.8284\gt2$ and the minimum of $F(x)$is equal to $2.087\gt2$ and the study of the curve can be stop here (by symmetry). Similarly with all other segment with $a\in[0.36,5.02]$ , the problem with $a\ne4$ is analogue. You can verify that for values of $a$ in $[1.77,2.3]$ the minimum of the function $F(x)$ is very near of $2$ but always greater that $2$.

It is clear that you can also calculate the minimum of $F(x)$ for a literal $a$ belonging to the interval above mentioned.