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$a,b,c \geqslant 0,$$ a+b+c=3$, and $(a+b)(b+c)(c+a) \neq 0$ , prove

$$\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$$

I try Bernouli's inequality but checking the case of $a>1$ and $0<a<1$ is truly complicated. I try to come up with some estimations (i.e., $x^x \geqslant \frac12 (1+x^2$)) but has not yet been successful.

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    $\begingroup$ Graphical and numerical evidence shows that the inequality is indeed true. But another answer in the same style doesn't contribute much. Still apart from the fact that I'd rather like calculus to be my top tag instead of inequality. $\endgroup$ Jun 29, 2016 at 10:11
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    $\begingroup$ Just my two cents: denote the given expression as $F \equiv f(a, b, c) \geq 3$. Consider the 3 fold symmetry in interchanging $a \leftrightarrow b$, $b \leftrightarrow c$, and $a \leftrightarrow c$. The three exchanges yield the same expression, which I shall denote as $G \equiv f(b, a, c) = f(a, c, b) = f(c, b, a)$. Given a set of fixed ${a, b, c}$ note that $G \neq F$ is a different quantity, yet as a whole $G$ is equivalent to $F$ so one might want to consider either $G \geq 3$ or $G + F \geq 6$. (perhaps $F$ and $G$ are the 2 outcomes of f(a, b, c) on permutation groups of ${a, b, c}$) $\endgroup$ Sep 3, 2016 at 18:43
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    $\begingroup$ New proof is presented.. $\endgroup$ Jul 2, 2018 at 13:31

2 Answers 2

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$$\mathbf{\color{green}{New\ proof,\ version\ of\ 02.07.18}}$$


$$\mathbf{\color{brown}{Using\ inequality}}$$

Let us consider the inequality $$(1+t)^{-\alpha}\geq1-\alpha t\quad \text{ for } t>-1,\quad \alpha>0,\tag1$$ which works due to the Maclauin series of $$(1+t)^{-\alpha} = 1 - \alpha t+\dfrac12(-\alpha)(-\alpha-1)t^2+\dfrac16(-\alpha)(-\alpha-1)(\alpha-2)t^3+\dots$$

Using of the inequality $(1)$ in the form of $$\left(\dfrac{a+1}{a+b}\right)^a=\left(\dfrac{a+b}{a+1}\right)^{-a}= \left(1+\dfrac{b-1}{a+1}\right)^{-a}\ge1-\dfrac{a(b-1)}{1+a}=2-b+\dfrac{b-1}{1+a}\tag2$$ is correct, because for the issue conditions $$a>0,\ b>0,\ c>0,\quad a+b+c = 3\tag3$$ $$\dfrac{b-1}{a+1} > -1.$$

$$\mathbf{\color{brown}{Task\ transformation}}$$

Using the inequality $(2)$ with the constraints $(3),$ easily to get $$\left(\dfrac{a+1}{a+b}\right)^a + \left(\dfrac{b+1}{b+c}\right)^b + \left(\dfrac{c+1}{c+a}\right)^c \geq 3+ \dfrac{b-1}{1+a}+ \dfrac{c-1}{1+b}+ \dfrac{a-1}{1+c},$$ so it can be proved the stronger inequality than the issue one: $$\dfrac{b-1}{1+a}+ \dfrac{c-1}{1+b}+ \dfrac{a-1}{1+c}\ge0,$$ or $$(b^2-1)(1+c)+(c^2-1)(1+a)+(a^2-1)(1+b)\ge0.\tag4$$ The least value of $LSH(4)$ can be achieved in the stationary points or on the edges of the area $(3).$

$$\mathbf{\color{brown}{Stationary\ points}}$$ The stationary points of the function can be found using the Lagrange multiplyers method, as the stationary points of the function $$f(a,b,c,\lambda)=(b^2-1)(1+c)+(c^2-1)(1+a)+(a^2-1)(1+b)+\lambda(a+b+c-3),$$ by the solving of the systen $f'_a = f'_b = f'_c = f'_\lambda=0,$ or \begin{cases} c^2-1+2a(1+b)+\lambda=0\\ a^2-1+2b(1+c)+\lambda=0\\ b^2-1+2c(1+a)+\lambda=0\\ a+b+c=3.\tag5 \end{cases} Summation of $(5.1)-(5.3)$ gives $\lambda=-4,$ then \begin{cases} c^2+2a(1+b)=a^2+2b(1+c)=b^2+2c(1+a)=5\\ a+b+c=3, \end{cases} \begin{cases} c^2+2a(4-a-c)=a^2+2(3-a-c)(1+c)=5\\ a+b+c=3, \end{cases} \begin{cases} 3c^2=3a^2-10a+4c+6\\ 3a^2-10a+4c+6+6a(4-a-c)=15\\ a+b+c=3, \end{cases} \begin{cases} (4-6a)c=3a^2-14a+9\\ 3(3a^2-14a+9)^2=(4-6a)^2(3a^2-10a+6)+4(4-6a)(3a^2-14a+9)\\ a+b+c=3, \end{cases} \begin{cases} (a-1)(27a^3-81a^2+45a+1)=0\\ c=\dfrac{3a^2-14a+9}{4-6a}\\ a+b+c=3,\tag6 \end{cases} The root $a=1$ leads to the solution $$a=b=c=1,\quad f(a,b,c,\lambda)=0.\tag7$$ Taking in account the task symmetry by the variables $a,b,c,$ the roots of cubic part of $(6.1)$ are the values of $a,b,c.$ At the same time, production of this roots is negative due to the Vieta theorem. This means that the solution $(7)$ is the single one, which satisfies conditions $(3).$


$$\mathbf{\color{brown}{The\ edges}}$$ To analyze the edges of area let us consider the case $a\to 0.$ It leads to the inequality in one variable $$(b^2-1)(2-b)+(3-b)^2-b-2\ge0,$$ or $$3-b(b-2)(b-3)\ge0.\tag8$$ Taking in account that for $b\in[2,3]$ $LSH(8)\ge3,$ and for $b\in(0,2)$ $$LSH(8)=3-b(2-b)(3-b)\ge3-\dfrac{(5-b)^3}9\ge0,$$ the inequality $(8)$ holds in the edges.

$$\mathbf{\color{green}{Proved}}$$


$$\mathbf{\color{black}{Old\ proof}}$$


$$\mathbf{\color{brown}{Task\ transformation}}$$

First, we should consider the inequality within the area. Using evident inequality $$\dfrac1{1-t}\geq1+t\quad \text{ for } t\in\left(-1,1\right)\qquad(1)$$ and AM-GM for $a,b,c>0,$ one can get: $$\dfrac{a+1}{a+b}=\dfrac{a+1}{3-c}=\dfrac12\,\dfrac{a+1}{1-\dfrac{c-1}2}\geq\dfrac{a+1}2\left(1+\dfrac{c-1}2\right) = \dfrac{a+1}2\,\dfrac{c+1}2\geq\sqrt{ac},$$ $$\left(\dfrac{a+1}{a+b}\right)^a + \left(\dfrac{b+1}{b+c}\right)^b + \left(\dfrac{c+1}{c+a}\right)^c \geq (ac)^{a/2} + (ba)^{b/2} +(cb)^{c/2}\\ =\exp\left(\frac{a}2\,\ln(ac)\right) + \exp\left(\frac{b}2\,\ln(ba)\right) + \exp\left(\frac{c}2\,\ln(cb)\right)\\ \geq 3\exp\left(\frac{a}6\,\ln(ac) + \frac{b}6\,\ln(ba)+\frac{c}6\,\ln(cb)\exp\right)\\ =3\exp\left(\frac{a+b}6\ln{a}+\frac{b+c}6\,\ln{b}+\frac{c+a}6\,\ln{c}\right) \\ =3\exp\left(\frac{3-c}6\,\ln{a}+\frac{3-a}6\,\ln{b}+\frac{3-b}6\, \ln{c}\right),$$ and that gives the possibility to prove the inequality $$(3-c)\ln{a}+(3-a)\ln{b}+(3-b)\ln{c}\geq0\tag2$$ for $a,b,c>0,\ a+b+c=3.$


$$\mathbf{\color{brown}{Stationary\ points}}$$ To do this, it suffices to find the least value of the function $$f(a,b) = (a+b)\ln{a}+(3-a)\ln{b}+(3-b)\ln(3-a-b).\tag3$$ The stationary points of the function can be found by equating to zero the partial derivatives $f'_a$ and $f'_b,$ or $$\begin{cases} \ln{a}+\dfrac{a+b}a -\ln{b} - \dfrac{3-b}{3-a-b} = 0\\[4pt] \ln{a}+\dfrac{3-a}b - \ln(3-a-b) - \dfrac{3-b}{3-a-b} =0, \end{cases}$$ $$\begin{cases} \ln{\dfrac{a}b} + \dfrac{a}{b} - \dfrac{a}{3-a-b} = 0\\[4pt] \ln{\dfrac{a}{3-a-b}} + \dfrac{3-a-b}{b} - \dfrac{a}{3-a-b} = 0,\\ \end{cases}$$ or $$\begin{cases} \ln{xy} - x + xy =0\\[4pt] \ln{x} -x + y =0\\[4pt] x=\dfrac{a}{3-a-b}\\[4pt] y=\dfrac{3-a-b}{b}, \end{cases}$$ $$\ln{y}+y(x-1)=0,\quad y=x-\ln{x},$$ $$\ln(x-\ln{x})+(x-1)(x-\ln{x})=0,\tag4$$ with single solution $$x=y=a=b=1,$$ which corresponds to the maximum $f(a,b)$ $$f_m=0.$$


$$\mathbf{\color{brown}{The\ edges}}$$ To analyze the edges of area let us consider the case $a=0.$ It leads to the inequality in

one variable $$\left(\dfrac{b+1}3\right)^b+\left(\dfrac{4-b}{3-b}\right)^{3-b}\geq 2,\tag5$$ which is satisfied for $0<b<3.$

Thus, $$\boxed{\left(\dfrac{a+1}{a+b}\right)^a + \left(\dfrac{b+1}{b+c}\right)^b + \left(\dfrac{c+1}{c+a}\right)^c \geq 3}.$$

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    $\begingroup$ I think your first line is incorrect : AM-GM yields $\frac{1}{a+b} \leq \frac{1}{2\sqrt{ab}}$, not $\frac{1}{a+b} \geq \frac{1}{2\sqrt{ab}}$ as you claim $\endgroup$ Aug 31, 2016 at 10:30
  • $\begingroup$ @EwanDelanoy Thank you, fixed $\endgroup$ Sep 3, 2016 at 11:05
  • $\begingroup$ @DanielBuck Thanks, the borders of area should be considered $\endgroup$ Sep 3, 2016 at 16:04
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    $\begingroup$ also, there is a missing factor $3$ when you use convexity of $\exp$. There is also one $\exp$ redundant (right after $\frac c6 \ln(ca)$). $\endgroup$
    – timon92
    Sep 3, 2016 at 16:58
  • $\begingroup$ @timon92 Thank you, typo is fixed $\endgroup$ Sep 3, 2016 at 17:09
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Yuri Negometyanov gave $$\left(\frac{a+1}{a+b}\right)^a = \left(1 + \frac{b-1}{a+1}\right)^{-a} \ge 1 - a\cdot \frac{b-1}{a+1} = 2 - b + \frac{b-1}{a+1}. \tag{1}$$ This follows from the Bernoulli inequality $(1+x)^r \ge 1 + rx$ for $x > -1$ and $r \le 0$.

Using (1), it suffices to prove that $$\frac{b-1}{1+a}+ \frac{c-1}{1+b}+ \frac{a-1}{1+c}\ge 0. \tag{2}$$

We can prove (2) in another way as follows.

After clearing the denominators, it suffices to prove that $$a^2b + b^2c + c^2a + a^2 + b^2 + c^2 - a - b - c - 3 \ge 0$$ or (since $a+b+c=3$) $$a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 \ge 0.$$ By Cauchy-Bunyakovsky-Schwarz inequality, we have $$a^2b + b^2c + c^2a \ge \frac{(ab + bc + ca)^2}{b + c + a} = \frac{(ab + bc + ca)^2}{3}.$$ Thus, we have \begin{align} a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 &\ge \frac{(ab + bc + ca)^2}{3} - 2(ab+bc+ca) + 3\\ &= \frac{1}{3}(ab + bc + ca - 3)^2\\ &\ge 0. \end{align}

We are done.

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