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Let $a,b,c >0$ with $a+b+c=3$. Prove that $$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52.$$

What I did:

It is cyclic inequality so I assume $c= \min\{ a,b,c \}$.
I consider the first case where $a\ge b\ge c$ then $$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$$ $$\Leftrightarrow \frac{a^a}{a^b} +\frac{b^b}{b^c}+\frac{c^a}{c^c}> \frac52$$ I check function $f(x) =x^x$ to see if it is a strictly monotonic function or not. It turns out that it is a concave up function so I get stuck here.

River Li
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HN_NH
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2 Answers2

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enter image description here

Not a bounty candidate. Just a pictorial comment.
Make an isoline/contour plot in the $(a,b)$-plane of the function: $$ f(a,b) = a^{|b-a|}+b^{|c-b|}+c^{|a-c|} - \frac52 \quad \mbox{with} \quad c=3-a-b $$ Then this is what we get. The blue spots are where $\,|f(a,b)| < 0.02$ . There seem to be several of these minimum values. I wish the rigorous proof producers among us good luck.

Han de Bruijn
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  • @zhoraster: Yes, oops, I've corrected it. – Han de Bruijn Feb 06 '17 at 17:52
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    The reason there are 3 minimums is because of the cyclic nature of the equation. If one rotates the values of a minimum solution $(a_1,b_1,c_1)$ to $(c_1, a_1, b_1)$ or to $(b_1,c_1, a_1)$, one gets the same equation. – Jens Feb 06 '17 at 17:59
  • Perhaps you can confirm the minimum solution I found, which was approx. $(0.19889, 1.59393, 1.20718)$, giving a value of $2.511731087$. – Jens Feb 06 '17 at 18:27
  • @Jens: Got something comparable, probably less accurate though: $f( 0.202, 1.207, 1.592) = 2.512$ with an error of $\approx 0.007$ . – Han de Bruijn Feb 07 '17 at 13:37
  • @Jens: If you can guarantee your results with the given accuracy, then perhaps I can do something with it. Can you? – Han de Bruijn Feb 07 '17 at 13:57
  • Well, in my answer above I was relying on $9$ decimal digit accuracy in Excel. I could also have given the values $(0.1989, 1.59393, 1.20717)$ as they gave the same result to $9$ digits. If I can trust $12$ digits of accuracy, I can squeeze in another digit in the answer giving $(0.1989, 1.593927, 1.207173)$ with value $2.511731086927$. – Jens Feb 07 '17 at 17:56
  • @Jens: Did some attempts with converting your values into exact fractions, while hoping to find an exact minimum. Without any success, though. – Han de Bruijn Feb 08 '17 at 12:45
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Perhaps it would help to take $a=1+x$, $b=1+y$, $c=1+z$, so that $x+y+z=0$

Now the inequality looks like:

$$(1+x)^{|y-x|} + (1+y)^{|z-y|} + (1+z)^{|x-z|} \geq \frac{5}{2}$$

Using bernoulli's inequality, the expression is more or less

$$1 + x|y-x| + 1 + y|z-y| + 1 + z|x-z| \geq \frac{5}{2}$$ $$x|y-x| + y|z-y| + z|x-z| \geq \frac{-1}{2}$$

And we can split it in two cases depending on how x, y and z are sorted.

For example, if $x<y<z$, we must prove that: $$x(y-x) + y(z-y) + z(z - x) = -x^2 - y^2 + z^2 + xy + yz - xz \geq \frac{-1}{2}$$

We recall that $x+y+z=0$, and thus $z = -x-y$:

$$-x^2 - y^2 + z^2 + xy + yz - xz = -x^2 - y^2 + x^2 + y^2 - 2xy + xy - (y+x)(y-x) = x^2 - y^2 - xy$$

It should not be hard to check if this is more that $\frac{-1}{2}$ given our restraints.

gonthalo
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    Your Bernoulli's inequality is only correct if |y-x| >1, and so on... which is not always true – HN_NH Feb 03 '17 at 02:31