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If $a,b,c,d >0$, and $a+b+c+d=4$, prove that $$a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi.$$

I don't think Jensen's inequality will help here, but I think first determining where equality holds will be useful. Maybe taking the logarithm or exponential of both sides will also be useful, but I want to in the end get rid of the plus signs in order to simplify it.

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    $\begingroup$ This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though... $\endgroup$ – Surb May 6 '16 at 20:54
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    $\begingroup$ May I ask how you came across this problem? Also, do you know whether the constant $\pi$ is optimal? $\endgroup$ – Wojowu May 6 '16 at 20:55
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    $\begingroup$ The best I've got out of Mathematica is 3.1605859174508652189…, using NMinimize[{(a^2)^(a^2 b^2) + (b^2)^(b^2 c^2) + (c^2)^(c^2 d^2) + \ (d^2)^(d^2 a^2), a != 0 && b != 0 && c != 0 && d != 0 && a^2 + b^2 + c^2 + d^2 == 4}, {a, b, c, d}, WorkingPrecision -> 100] $\endgroup$ – Patrick Stevens May 6 '16 at 21:26
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    $\begingroup$ Alright, did I fall for a "scam", i.e. someone cooked up a function and figured out a numeric lower bound? I found the original post on artofproblemsolving and it seems the original poster was banned from the site. Is there an actual solution to this problem? $\endgroup$ – Ivan Mar 14 '17 at 2:39
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    $\begingroup$ An interesting inequality, but even without the numerical evidence for the bound being around $3.16$ there doesn't seem to be any reason for $\pi$ to feature here. Seems arbitrary $\endgroup$ – Yuriy S Apr 25 '18 at 10:04
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TL;DR: The inequality has been proven for all cases except the following five:

  • $1 < a < 2$, $b < 1$, $c > 1$, $d < 1$

  • $1 < a < 2$, $b < 1$, $c < 1$, $d < 1$

  • $2 < a < 3$, $b < 0.207$, $c > 1$, $d < 1$

  • $2 < a < 3$, $b < 0.256$, $c < 1$, $d < 1$

  • $3 < a < 4$, $b < 0.129$, $c < 1$, $d < 1$


This partial answer heavily uses the results that for a real number $k>0$,

  • $\min x^{kx}=(\sqrt[e]{e^k})^{-1}$,

  • $k^{kx}>k^{kx_0}$ for $k>1$ and $x>x_0$,

  • $k^{kx}<k^{kx_0}$ for $k<1$ and $x>x_0$.

As the sum $$S=a^{ab}+b^{bc}+c^{cd}+d^{da}$$ is cyclic, we need only concern $a=1$ and $a>1$ to prove the truth of the inequality. Thus there are sixteen cases that we need to consider (note that in most cases the condition $a+b+c+d=4$ is implicitly used).

$1)$ $a=b=c=d=1$

Clearly $S=1+1+1+1>\pi$.

$2)$ $a=b=1$, $c>1$, $d<1$

As $c<2$, $S>1+1+1+(\sqrt[e]{e})^{-1}>\pi$.

$3)$ $a=b=1$, $c<1$, $d>1$

As $d<2$, $S>1+1+(\sqrt[e]{e^2})^{-1}+1>\pi$.

$4)$ $a=1$, $b>1$, $c>1$, $d<1$

We have $S>1+1+1+(\sqrt[e]{e})^{-1}>\pi$.

$5)$ $a=1$, $b>1$, $c<1$, $d>1$

As $d<2$, $S>1+1+(\sqrt[e]{e^2})^{-1}+1>\pi$.

$6)$ $a=1$, $b>1$, $c<1$, $d<1$

We have $S>1+1+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e})^{-1}>\pi$.

$7)$ $a=1$, $b<1$, $c>1$, $d<1$

If $b\ge0.6$, $c\le2.4$ so $S\ge1+0.6^{0.6\cdot2.4}+1+(\sqrt[e]{e})^{-1}>\pi$. If $b<0.6$, $c>1.4$ so $S>1+(\sqrt[e]{e^3})^{-1}+\min\{1.4^{1.4d}+d^d\}>\pi$.

$8)$ $a=1$, $b<1$, $c<1$, $d>1$

If $b\ge0.89$, $d\le2.11$ so $S\ge1+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^{2.11}})^{-1}+1>\pi$. If $b<0.89$, $3>d>1.11$ so $S\ge 1+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^3})^{-1}+1.11^{1.11}>\pi$.

$9)$ $a=1$, $b<1$, $c>1$, $d>1$

As $c<2$, $S>1+(\sqrt[e]{e^2})^{-1}+1+1>\pi$.

$10)$ $a>1$, $b>1$, $c>1$, $d<1$

As $a<2$, $S>1+1+1+(\sqrt[e]{e^2})^{-1}>\pi$.

$11)$ $a>1$, $b>1$, $c<1$, $d>1$

As $d<2$, $S>1+1+(\sqrt[e]{e^2})^{-1}+1>\pi$.

$12)$ $a>1$, $b>1$, $c<1$, $d<1$

If $d\ge0.675$, $S>1+1+(\sqrt[e]{e^{0.675}})^{-1}+0.675^{0.675}>\pi$. If $a\ge2$, $d<0.675$, then $S>2^2+0+0+0>\pi$. If $a<2$, $d<0.675$, then $S>1+1+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^2})^{-1}>\pi$.

$13)$ $a>1$, $b<1$, $c>1$, $d<1$

If $3>a>2$, $c<2$ so $b\ge0.207$, $S>2^{2\cdot0.207}+(\sqrt[e]{e^2})^{-1}+1+(\sqrt[e]{e^3})^{-1}>\pi$.

$14)$ $a>1$, $b<1$, $c<1$, $d>1$

If $d\ge2$, $S>0+0+0+2^2>\pi$. If $d<2$, $S>1+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^2})^{-1}+1>\pi$.

$15)$ $a>1$, $b<1$, $c>1$, $d>1$

As $c<2$, $S>1+(\sqrt[e]{e^2})^{-1}+1+1>\pi$.

$16)$ $a>1$, $b<1$, $c<1$, $d<1$

If $4>a>3$, $b\ge0.129$, $S>3^{3\cdot0.129}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^4})^{-1}>\pi$. If $3>a>2$, $b\ge0.256$, $S>2^{2\cdot0.256}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^3})^{-1}>\pi$.

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  • $\begingroup$ There miss many cases by example when $a<1$ and we don't have the case when the minimum is reached... $\endgroup$ – user635269 Apr 19 at 13:59
  • $\begingroup$ @FatsWallers As I've said in my post, because the sum $a^{ab}+b^{bc}+c^{cd}+d^{da}$ is cyclic, proving the inequality for $a\ge 1$ is sufficient due to the constraint $a+b+c+d=4$. If you try doing it for $a<1$, you get the same conclusions as for say, $b<1$ or $c<1$. $\endgroup$ – TheSimpliFire Apr 19 at 15:41
  • $\begingroup$ Okay sorry I read too fast lol ^^ . $\endgroup$ – user635269 Apr 20 at 12:44
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We can easily prove a weaker inequality as follows. It is known that if $x>0$ then $x^x\ge \left(\tfrac 1e\right)^{1/e}=E$. Since the original inequality is cyclic, witout loss of generality we can assume that $d\ge 1$. Then

$$a^{ab}+b^{bc}+c^{cd}+d^{da}\ge E^b+E^c+E^d+1^a\ge 3E^{(b+c+d)/3}+1\ge 3E^{4/3}+1>2.8369.$$

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Warning : It's a partial proof but I think it's interesting . We have the following theorem :

Let $a,b,c,d>0$ such that $abcd\geq e^{-2}$ then we have : $$a^{ab}+b^{bc}+c^{cd}+d^{da}\geq 4(e^{-0.5})^{e^{-1}}$$

Proof : We begin to apply AM-GM to get : $$a^{ab}+b^{bc}+c^{cd}+d^{da}\geq 4(a^{ab}b^{bc}c^{cd}d^{da})^{0.25}$$ Hence we want to prove : $$4(a^{ab}b^{bc}c^{cd}d^{da})^{0.25}\geq 4(e^{-0.5})^{e^{-1}}$$ Now we take the logarithm : $$0.25\sum_{cyc}ab\ln(a)\geq -0.5e^{-1}$$

But $x\ln(x)$ is convex so we get : $$0.25\sum_{cyc}ab\ln(a)\geq 0.25(ab+bc+cd+da)\ln(\frac{ab+bc+cd+da}{a+b+c+d})$$

Hence we want to prove :

$$0.25(ab+bc+cd+da)\ln(\frac{ab+bc+cd+da}{a+b+c+d})\geq -0.5e^{-1}$$

Wich is true because : $$\frac{ab+bc+cd+da}{a+b+c+d}\geq e^{-0.5}$$

And

$$a+b+c+d\geq 4e^{-1}$$

Now remains to add the condition $a+b+c+d=4$ to get :

Let $a,b,c,d>0$ such that $abcd\geq e^{-2}$ and $a+b+c+d=4$ then we have : $$a^{ab}+b^{bc}+c^{cd}+d^{da}> 4(e^{-0.5})^{e^{-1}}$$

An other idea will be to extend the initial theorem (using the same technique) and see what happend .

Finally we conclude with the fact that it completes partialy the result of TheSimpliFire .

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    $\begingroup$ Why $\frac{ab+bc+cd+da}{a+b+c+d}\geq e^{-0.5}$? Why this and $a+b+c+d\geq 4e^{-1}$ (by the way, since $abcd\ge e^-2$ then AM-GM implies $a+b+c+d\geq 4e^{-1/2}$) imply $0.25(ab+bc+cd+da)\ln(\frac{ab+bc+cd+da}{a+b+c+d})\geq -0.5e^{-1}$? Note, that a priori the logarithm may be negative, so we have to take this into account. $\endgroup$ – Alex Ravsky May 1 at 17:01

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