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

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 $$>1$$

• We can rewrite as $A^{B^2+1}+B^{A^2+1}\ge2AB$ which is similar to this problem. Oct 9, 2021 at 10:42
• @RounakSarkar I believe there's sufficient context. They have linked to a relevant inequality, shown an attempt and identified the particular issue in their attempt. Oct 13, 2021 at 15:24
• @RounakSarkar Interest isn't enough to keep a question up on its own for sure, but you'll slowly get a hang of how context checking works in the contest/inequality setting. Since an attempt is quite difficult to provide, "source" and "linked-questions" are the two preferred ways of adding context, and there is a (somewhat) useful linked question here, and there is an attempt (although I'd put a source above that, but the attempt shows that the question isn't too trivial to admit AM-GM immediately). So this is ok for a contest/inequality type problem. Oct 14, 2021 at 5:18
• This seems to be a hard nut. Some remarks: If $a, b \ge 1$ or $a, b \le 1$ then $a^{b-1}b^{a-1} \ge 1$ and the conclusion follows (that are the easy cases). If $0 < a \le 1/4$ and $b > 1$ then $\sqrt{b^a/a} \ge \sqrt{1/a} \ge 2$. If $1/4 < a < 1$ and $b \ge 256$ then $\sqrt{b^a/a} \ge \sqrt{b^a} \ge 2$. – It remains “only” the rectangle $1/4 < a < 1$, $1 < b < 256$ to investigate. Oct 14, 2021 at 6:56
• @MartinR We can do better :Note that if either $\sqrt{\frac{a^b}{b}} \geq 2$ or $\sqrt{\frac{b^a}{a}} \geq 2$ then we are done. The former case translates to $a \geq (4b)^{\frac 1b}$ and the latter to $b$ switched with $a$. The function $(4b)^{\frac 1b}$ attains a maximum of $e^{\frac 4e} \approx 4.35$ over $b>0$, therefore if either $b$ or $a$ is above $4.35$ then the inequality holds because that particular term on the LHS is above $2$. So this further restricts the domain of investigation. Oct 14, 2021 at 7:03

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.

$$\phantom{2}$$

Update: I found a simpler proof.

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 $$0 < y \le 1 \le x$$, $$\frac{x^{y^2}}{y} + \frac{y^{x^2}}{x} \ge 2.$$

$$\phantom{2}$$

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.)

$$\phantom{2}$$

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.

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.

• Good solution! It is very useful Oct 14, 2021 at 22:44
• @3017 Thanks. Honestly, it is not nice. :) Oct 15, 2021 at 2:00
• Nice!+1,can you explain How to found the fact 1 and fact 2? Oct 15, 2021 at 6:34
• @functionsug It is just (1,1)-Pade approximant of $x^{y^2}$ at $x = 1$. Oct 15, 2021 at 6:57
• @River Li.-Good your answer to this defiant question. Congratulations. It would be even easier if we consider for all point $(a,b)$ in the first quadrant the following function of $[a,b]$ (or $[b,a]$ to have a sens) in $\mathbb R$ $$F(x)=\sqrt{\frac{x^{s-x}}{s-x}}+\sqrt{\frac{(s-x)^x}{x}}$$ and determine its minimum. This minimum is always greater than 2 and is the smallest possible value for all point $(a,b)$ in the line $x+y=s$ Unfortunately this minimum should be calculated using numerical methods, which is contrary to the local maximun equal to $F(\frac s2)$. (suite) Oct 16, 2021 at 16:44

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

• That are the two simple cases. But what if $0 < a < 1 < b$ or $0< b < 1 < a$? Oct 13, 2021 at 19:33
• @MartinR you are right! I will try to deal with those cases and see if I get something Oct 13, 2021 at 20:05
• It is already noted in the question that $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2\left(a^{b-1}b^{a-1}\right)^{1/4}$, and it is easy to see that $a^{b-1}b^{a-1} \ge 1$ if $a, b$ are both $\ge 1$ or both $\le 1$. That's why I called these the “simple cases.” Oct 14, 2021 at 6:40
• @MartinR I did not notice that, good point. Oct 14, 2021 at 6:54

Alternative sketch of proof for $$0

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 !

• @RiverLi What do you think about ? Thanks ! Jan 31, 2022 at 13:45
• It is true. Your proof is better than mine. Very nice! Apr 1, 2022 at 1:07
• @RiverLi Thanks you I spend some hours on my computer for that . Glad you like it . Thanks you again ! Apr 5, 2022 at 8:38
• You know, this time, you gave rational bounds for $x^{y^2}$. Apr 5, 2022 at 11:19

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.

$$\phantom{2}$$

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.

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.

• “You can verify that... the minimum of the function $F(x)$ is very near of $2$ but always greater that $2$.” – I miss a rigorous proof here. Oct 16, 2021 at 1:00
• @Martin R.- I agree with you (I gave this for anyone who wanted to verify it). Read my comment to River Li's anwer, please. It is an answer more concise to the question then the mine here. Oct 16, 2021 at 16:54

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

• “partial answer”, “really easy”, “It seems we have” – sorry, but all that sounds quite vague to me. Can you clearly state which cases you have proved, and what is just a conjecture? Oct 9, 2021 at 10:58
• How do you conclude from $h(2-x)+h(x)\ge 4$ that $h(x) \ge 2$? We don't have $h(2-x) =h(x)$, so what symmetry are you referring to? Oct 11, 2021 at 12:22
• @MartinR Oh well to the line $x=1$ on $(0,2)$ I add it ? Oct 11, 2021 at 14:05
• @ErikSatie: But that is wrong. $h$ is not symmetric to $x=1$. $h(2-x) = h(x)$ is not true. Oct 11, 2021 at 14:19
• @Eudoxus Perhaps a conflict of interest ...Or people vote for bad thing sometimes Oct 12, 2021 at 14:06