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

Question

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.

Answer 1

TL;DR: The inequality has been proven for all cases except the following five:

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

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

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

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

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

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

Answer 2

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 .

Answer 3

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

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

Answer 4

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.