# Prove that $a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi$

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.

• This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though...
– Surb
May 6, 2016 at 20:54
• May I ask how you came across this problem? Also, do you know whether the constant $\pi$ is optimal? May 6, 2016 at 20:55
• 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] May 6, 2016 at 21:26
• 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?
– Ivan
Mar 14, 2017 at 2:39
• 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 Apr 25, 2018 at 10:04

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 $$x>x_0$$,

• $$k^{kx} for $$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 $$3>a>2$$, $$b\ge0.256$$, $$S>2^{2\cdot0.256}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^3})^{-1}>\pi$$.

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 .

• 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. May 1, 2019 at 17:01

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

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

With computer, we can use the branch and bound strategy to prove the inequality.

Proof: The inequality is written as $$F(a, b, c, d) = \mathrm{e}^{ba\ln a} + \mathrm{e}^{cb\ln b} + \mathrm{e}^{dc\ln c} + \mathrm{e}^{a d\ln d} > \pi.$$

We first introduce some auxiliary results (Facts 1 through 2). The proofs are given later.

Fact 1: Given four real numbers $$0 \le A < B, \ 0 \le C < D$$ with $$B-A \le \frac{1}{8}, \ D-C\le \frac{1}{8}$$. Let $$x\in [A, B]$$ and $$y\in [C, D]$$. For convenience, we set $$0\ln 0 = 0$$. We have

i) If $$\frac{1}{\mathrm{e}} \in [A, B]$$, then $$yx\ln x \ge -D \mathrm{e}^{-1}$$;

ii) If $$\frac{1}{\mathrm{e}} \notin [A, B]$$ and $$A < 1$$, then $$yx\ln x \ge D \min(A\ln A, B\ln B)$$;

iii) If $$A \ge 1$$, then $$yx\ln x \ge C A \ln A$$.

Fact 2: If $$a\in [0, 1], \ b \in [0, 2], \ c \in [0, 1], \ d\ge 0$$ and $$a+b+c+d=4$$, then $$F(a,b,c,d) > \pi$$.

Now let us proceed. WLOG, assume that $$d= \max(a,b,c,d)$$. Clearly $$b \le 2$$.

If $$a > 1$$, we have $$\mathrm{e}^{ba\ln a} + \mathrm{e}^{cb\ln b} + \mathrm{e}^{dc\ln c} + \mathrm{e}^{a d\ln d} \ge 2\mathrm{e}^{\frac{1}{2}(cb\ln b + dc\ln c)} + 2 \ge 2\mathrm{e}^{\frac{1}{2}(-\frac{3}{\mathrm{e}})} + 2 > \pi$$ since $$x\ln x \ge -\frac{1}{\mathrm{e}}$$ for all $$x > 0$$, and $$x\mapsto \mathrm{e}^x$$ is convex.

If $$c > 1$$, we have $$\mathrm{e}^{ba\ln a} + \mathrm{e}^{cb\ln b} + \mathrm{e}^{dc\ln c} + \mathrm{e}^{a d\ln d} \ge 2\mathrm{e}^{\frac{1}{2}(ba\ln a + cb\ln b)} + 2 \ge 2\mathrm{e}^{\frac{1}{2}(-\frac{b+c}{\mathrm{e}})} + 2 \ge 2\mathrm{e}^{\frac{1}{2}(-\frac{8/3}{\mathrm{e}})} + 2 > \pi.$$

If $$a \le 1$$ and $$c\le 1$$, from Fact 2, the desired result follows.

We are done.

$$\phantom{2}$$

Proof of Fact 1: Let $$f(x) = x\ln x$$. Then $$f(x)$$ is strictly decreasing on $$[0, \frac{1}{\mathrm{e}})$$ and strictly increasing on $$(\frac{1}{\mathrm{e}}, \infty)$$. Also $$f(x)$$ achieves its global minimum at $$x = \frac{1}{\mathrm{e}}$$. It is easy to prove the desired results.

Proof of Fact 2: We use the branch and bound strategy.

Let $$\Omega = \{(a,b,c,d): \ a\in [0, 1], \ b \in [0, 2], \ c \in [0, 1], \ d\ge 0, \ a+b+c+d=4\}.$$ For $$i=0, 1, \cdots, 127; \ j=0, 1, \cdots, 255; \ k = 0, 1, \cdots, 127$$, let \begin{align} S(i,j,k) &= \Big\{(a,b,c,d): \ a \in \Big[\frac{i}{128}, \ \frac{i+1}{128}\Big], \quad b \in \Big[\frac{j}{128}, \ \frac{j+1}{128}\Big], \\ &\qquad\quad c \in \Big[\frac{k}{128}, \ \frac{k+1}{128}\Big], \quad d \in \Big[4 - \frac{3 + i + j + k}{128}, \ 4 - \frac{i + j + k}{128}\Big]\Big\}. \end{align} We have $$\Omega \subset \bigcup_{i,j,k} S(i,j,k)$$.

From Fact 1, we evaluate a lower bound $$L(i,j,k)$$ of $$F(a,b,c,d)$$ on $$S(i,j,k)$$, i.e., $$F(a,b,c,d) \ge L(i,j,k), \quad \forall (a,b,c,d) \in S(i,j,k).$$ Thus, we have $$F(a,b,c,d) \ge \min_{i,j,k} L(i,j,k), \quad \forall (a,b,c,d)\in \Omega.$$ With computer, we obtain $$\min_{i,j,k} L(i,j,k) > \pi$$. The desired result follows. This completes the proof of Fact 2.