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

Solution 1:

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

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

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

Solution 2:

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.