22
$\begingroup$

$a,b,c >0$, and $a+b+c=3$, prove $$ a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt[6]{5}$$

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

$\endgroup$
  • $\begingroup$ Did you get an answer for this problem ? I'm interested in this problem :) $\endgroup$ – pinkpanther5 Jun 8 '16 at 16:17
  • $\begingroup$ The problem is correct. I don't have a solution. $\endgroup$ – HN_NH Jun 8 '16 at 22:32
2
$\begingroup$

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 .

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

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 $$\sqrt[6]{5} = 1.30...$$ Seems like your inequality is true. Equality is when $x=0$.


12

$\endgroup$
  • $\begingroup$ Somehow, wolframalpha is not plotting this correctly. $f(3)=3$ but graph says $f(3)=0$. link $\endgroup$ – guest Feb 17 '17 at 1:17
  • $\begingroup$ 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 $x$, $0.261073$? $\endgroup$ – guest Feb 17 '17 at 1:48
0
$\begingroup$

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 $\,1.309… > 1.308 > \sqrt[6]{5}\,$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.