# Prove $a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt{5}$

$a,b,c >0$, and $a+b+c=3$, prove $$a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt{5}$$

I try to substitute $c=3-a-b$ to reduce the number of variables, but cannot further proceed to solve the problem. I made an Excel spreadsheet and test 100 pairs of $(a,b,c)$, it seems that the inequality is correct. I cannot even find where the equality occurs. Please help. This is a very unconventional problem

• Did you get an answer for this problem ? I'm interested in this problem :) – pinkpanther5 Jun 8 '16 at 16:17
• The problem is correct. I don't have a solution. – HN_NH Jun 8 '16 at 22:32

Hint: for $x>0$ and $y>0$ and $z>0$ we have : $$x+y+z\geq \left(\frac{x+1}{y+1}+\frac{y+1}{x+1}+\frac{x+1}{z+1}+\frac{z+1}{x+1}+\frac{z+1}{y+1}+\frac{y+1}{z+1}-1\right)^{\frac{1}{6}}\geq (5)^{\frac{1}{6}}$$ Method: If we put $x+y+z=\lambda$

with $x=a^{ab}a$

$y=b^{bc}b$

$z=c^{ac}c$

And $a+b+c=3$

The question is : when is the maximum reached in the following expression? $$\frac{x+1}{y+1}+\frac{y+1}{x+1}+\frac{x+1}{z+1}+\frac{z+1}{x+1}+\frac{z+1}{y+1}+\frac{y+1}{z+1}-1$$ The maximum is reached when $x=y=0$ and $z=\lambda$

Following this we have this inequality : $$\lambda\geq \left(1+2(\lambda+1+\frac{1}{1+\lambda})\right)^{1/6}$$ It occurs for $\lambda\simeq 1.36897$

Now the idea is to repeat the same reasoning with the following expression : $$\frac{x+2}{y+2}+\frac{y+2}{x+2}+\frac{x+2}{z+2}+\frac{z+2}{x+2}+\frac{z+2}{y+2}+\frac{y+2}{z+2}-1$$ We obtain this : $$\lambda\geq \left(1+2(\frac{2+\lambda}{2}+\frac{2}{2+\lambda})\right)^{\frac{1}{6}}$$ Its occurs for $\lambda\simeq 1.32985$

Now the idea is to take the following expression :

$$\frac{x+n}{y+n}+\frac{y+n}{x+n}+\frac{x+n}{z+n}+\frac{z+n}{x+n}+\frac{z+n}{y+n}+\frac{y+n}{z+n}-1$$ And with the same reasoning we obtain :

$$\lambda\geq \left(1+2(\frac{n+\lambda}{n}+\frac{n}{n+\lambda})\right)^{\frac{1}{6}}$$

We take the limit :

$$\lambda\geq \lim\limits_{n \to \infty}\left(1+2(\frac{n+\lambda}{n}+\frac{n}{n+\lambda})\right)^{\frac{1}{6}}$$ And we obtain $$\lambda \geq (5)^{\frac{1}{6}}$$

If I'm wrong tell me quickly .

• Could you explain why your hint is true and also how it relates to the OP question? – Jens Feb 14 '17 at 21:26
• So you mean, for $x>0$, $y>0$ and $z>0$, $x+y+z\geq (5)^{\frac{1}{6}}$? How about $x=y=z=0.1$? – guest Feb 15 '17 at 2:03
• if you take $z=a^{ab}b$,$y=b^{bc}c$,$z=a^{ac}a$ it works.With the following condition a+b+c=3 . – max8128 Feb 15 '17 at 3:48
• You might be right, but why? – timon92 Feb 15 '17 at 21:51
• I prepare an another proof . – max8128 Feb 16 '17 at 14:28

It is not an answer just a picture. I draw function $$f(x, y) = x^{xy}y + y^{yz}z + z^{zx}x,\quad z = 3 -x-y,\quad x,y\in[0,1.4).$$ Note that $$\sqrt{5} = 1.30...$$ Seems like your inequality is true. Equality is when $x=0$.

• Somehow, wolframalpha is not plotting this correctly. $f(3)=3$ but graph says $f(3)=0$. link – guest Feb 17 '17 at 1:17
• Minimize $f(x)$ over (0,3) works correctly. wolframalpha.com/input/?i=minimize+(x%2Bx%5E%7Bx(3-x)%7D*(3-x))+x%3D(0,3) Is there any closed form minimal $x$, $0.261073$? – guest Feb 17 '17 at 1:48

The equation system resulting from the problem situation cannot be solved elementarily, only numerically.

Using WolframAlpha

$$\text{minimize (y*x^(x*y) + (3-x-y)*y^(y*(3-x-y)) + x*(3-x-y)^(x*(3-x-y))) , x>0 , y>0 , x+y<3}$$

we get

$$\displaystyle \min(yx^{xy}+(3-x-y)y^{y(3-x-y)}+x(3-x-y)^{x(3-x-y)}\,|$$

$$\displaystyle\hspace{1cm} \,x>0\land y>0\land x+y<3)\approx 1.30948\,\,$$ at $$\,(x,y)\approx (0.261073,2.73893)\,$$

and it’s $$\,1.309… > 1.308 > \sqrt{5}\,$$ .