Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$

Question

$a,b,c >0$ and $a+b+c=3$, prove $$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$

I try to apply AM-GM
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3\cdot \sqrt[3]{\left(\frac{a+1}{a+b} \right)^{\frac25}\left(\frac{b+1}{b+c} \right)^{\frac25}\left(\frac{c+1}{c+a} \right)^{\frac25}}$$ Thus it remains to prove $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1 $$ with the condition $a+b+c=3.$
But I found the counter example for $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1 $$ :(

Answer 1

Only a partial answer.

Assuming $a\le b\le c$, then we have $0<a\le1$ and $1\le c<3$. Now we have two cases, $b\le1$ and $b>1$. The case $b\le1$ is easy to deal with.

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Assuming $0<a\le b\le1\le c<3$, we have \begin{align} \left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \ge 1 &\iff (a+1)(b+1)(4-a-b)\ge(a+b)(3-a)(3-b)\\ &\iff {\left(2-a-b\right)} {\left(1-a\right)} {\left(1-b\right)}\ge0. \end{align}

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Now assuming $0<a\le1\le b \le c<3$. Below is not an answer but only an analysis.

We want to know why this case is difficult, and why $\frac25$ is important.

Using the series expansion of $\exp$, and let $A = \operatorname{diag}\left(\ln\left(\frac{a+1}{a+b} \right),\ln\left(\frac{b+1}{b+c}\right),\ln\left(\frac{c+1}{c+a} \right)\right)$, then we have \begin{align} LHS &= \sum_{n=0}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!}\\ &= 3 + \frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3 +R_3, \end{align} where $R_3 = \sum_{n=4}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!} \ge 0$, since $e^x-(1+x+x^2/2+x^3/6)$ is positive for any $x\in\mathbb R$.

Then we can prove the inequality if we have $$\frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3\ge\!\!\!?\;0,$$ which can be simplified to $$75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge\!\!\!?\;0.\tag{1}$$

Numerical results suggest that using the 3 first terms is enough to prove the inequality. Note that in the first case where $b\le1$, using the first term $\operatorname{tr}A$ is enough (and what we did in the first part is in fact proving $\operatorname{tr}A\ge0$), that's why that case is easy.

So,

• Why the case $b\ge1$ is difficult?

  Because we have 2 more terms, $\operatorname{tr}A^2$ and $\operatorname{tr}A^3$, to deal with.

• Why $\frac25$ is important?

  Because $\frac25$ gives the coefficients 75, 15, and 2, which makes $75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge0$.

Answer 2

$\color{green}{\textbf{The bounds.}}$

For $c=0,\ b=3-a$ and WLOG $0<a\le\frac32,$ we have the inequality $$F(a) = \left(\frac{a+1}3\right)^{2/5} + \left(\frac{4-a}{3-a}\right)^{2/5} +\left(\frac{1}{a}\right)^{2/5}\geq3,$$ $$F'(a)=\frac25\left(\frac13\left(\frac{a+1}3\right)^{-3/5} + \frac1{(3-a)^2}\left(\frac{4-a}{3-a}\right)^{-3/5} - \left(\frac{1}{a}\right)^{7/5}\right),$$ with the root at $a_m\approx1.34387,$ so minimum of LHS achieves at $a=a_m$ and equals to $$3.00248>3,$$ then the inequality is satisfied.

The same result can be obtained, if to use the numerical inequalities

• $\frac{123}{136}>\left(\frac79\right)^{2/5}>\frac{104}{115},\; \frac{43}{57}>\left(\frac85\right)^{-3/5}>\frac{89}{118},\;\frac{123}{184}>\left(\frac34\right)^{7/5}>\frac{125}{187}.$

Then $$\lim\limits_{a\to 0}=-\infty,\quad F'\left(1\right) = \dfrac2{15}\left(\sqrt[\Large5]{\dfrac{27}8}+\dfrac34\sqrt[\Large5]{\dfrac8{27}}-3\right)< 0,$$ $$F'\left(\dfrac32\right) = \dfrac2{15}\left(\sqrt[\Large5]{\dfrac{125}{216}}+\dfrac43\sqrt[\Large5]{\dfrac{125}{27}}-2\sqrt[\Large5]{\dfrac23}\right) ,$$ $$F'\left(\dfrac32\right) = \dfrac2{15}\sqrt[\Large5]{\dfrac23}\left(\sqrt[\Large5]{\dfrac{125}{144}}+\dfrac43\sqrt[\Large5]{\dfrac{125}{18}}-2\right) >0,$$ $$F'\left(\dfrac43\right) =\frac25\left(\frac37\left(\frac79\right)^{2/5} + \frac9{25}\left(\frac85\right)^{-3/5} - \left(\frac34\right)^{7/5}\right) $$ $$<\frac25\left(\frac37\frac{123}{136}+\frac9{25}\frac{43}{57}-\frac{125}{187}\right) = -\dfrac{2711}{731500} < 0,$$ $$F'\left(\dfrac43\right) > \frac25\left(\frac37\frac{104}{115} +\frac9{25}\frac{89}{118}-\frac{123}{184}\right) = -\dfrac{17811}{4749500},$$

and for $a\in\left(\dfrac43,\dfrac32\right)$ $$F(a) > F\left(\dfrac43\right) + F'\left(\dfrac43\right)\left(a-\dfrac43\right)$$ $$=\left(\frac79\right)^{2/5} + \frac85\left(\frac85\right)^{-3/5} + \dfrac43\left(\frac34\right)^{7/5} + \dfrac16 F'\left(\dfrac43\right) \left(a-\dfrac43\right)$$ $$>\frac{104}{115}\left(1+\dfrac{6}{35}\left(a-\dfrac43\right)\right) +\frac{356}{295}\left(1+\dfrac9{100}\left(a-\dfrac43\right)\right) +\frac{500}{561} -\dfrac{123}{460}\left(a-\dfrac43\right),$$

$$F(a) > F\left(\dfrac43\right) + \dfrac16 F'\left(\dfrac43\right) $$ $$>\frac{104}{115}\,\frac{36}{35}+\frac{356}{295}\,\frac{203}{200}+\frac{500}{561} - \frac{41}{920} > 3.001768 >3.$$

$\color{green}{\textbf{Inequality transformation.}}$

Using the condition, one can write the original inequality in the form $$\left(\frac{a+1}{3-c}\right)^{\frac25}+\left(\frac{b+1}{3-a} \right)^{\frac25}+\left(\frac{c+1}{3-b}\right)^{\frac25}.\qquad(1)$$ Let $$x=\dfrac{3-a}{3-c},\quad y=\dfrac{3-b}{3-c},\qquad(2)$$ then $$\frac{-x+2y+2}3 = \frac{a+1}{3-c},\quad \frac{2x-y+2}{3x} = \frac{b+1}{3-a},\quad \frac{2x+2y-1}{3y}= \frac{c+1}{3-b}.\qquad(3)$$ For example, $$\dfrac{-x+2y+2}{3} = \dfrac{a-3+2(6-b-c)}{3(3-c)} = \dfrac{a-3+2(3+a)}{3(3-c)}=\dfrac{a+1}{3-c}.$$ This allows to prove the inequality $$\left(\frac{-x+2y+2}3\right)^{2/5}+\left(\frac{2x-y+2}{3x} \right)^{2/5}+\left(\frac{2x+2y-1}{3y}\right)^{2/5}\geq3$$ for $$x,y>0.$$

$\color{green}{\textbf{Primary optimization.}}$

To do this, it suffices to find the least value of the function $$f(x,y)=\left(\frac{-x+2y+2}3\right)^{2/5}+\left(\frac{2x-y+2}{3x} \right)^{2/5}+\left(\frac{2x+2y-1}{3y}\right)^{2/5}$$ for positive $x,y.$

The stationary points of the function can be found by equating to zero the partial derivatives $f_x$ and $f_y$, and that gives $$\begin{cases} -\dfrac25\left(\dfrac{-x+2y+2}1\right)^{-3/5}+\dfrac{2(y-2)}{5x^2}\left(\dfrac{2x-y+2}{x} \right)^{-3/5}+\dfrac4{5y}\left(\dfrac{2x+2y-1}{y}\right)^{-3/5}=0\\[4pt] \dfrac45\left(\dfrac{-x+2y+2}1\right)^{-3/5}-\dfrac2{5x}\left(\dfrac{2x-y+2}{x} \right)^{-3/5}-\dfrac{2(2x-1)}{5y^2}\left(\dfrac{2x+2y-1}{y}\right)^{-3/5}=0, \end{cases}$$ $$\begin{cases} (y-2)x^{-7/5}(2x-y+2)^{-3/5}+2y^{-2/5}(2x+2y-1)^{-3/5}=(-x+2y+2)^{-3/5}\\[4pt] x^{-2/5}(2x-y+2)^{-3/5}+(2x-1)y^{-7/5}(2x+2y-1)^{-3/5}=2(-x+2y+2)^{-3/5}, \end{cases}$$ $$ \begin{pmatrix}y-2&2y\\x&2x-1\end{pmatrix} \begin{pmatrix} x^{-7/5}\left(\dfrac{-x+2y+2}{2x-y+2}\right)^{3/5}\\ y^{-7/5}\left(\dfrac{-x+2y+2}{2x+2y-1}\right)^{3/5} \end{pmatrix} =\begin{pmatrix}1\\2\end{pmatrix}. $$ from whence $$ \begin{pmatrix} x^{-7/5}\left(\dfrac{-x+2y+2}{2x-y+2}\right)^{3/5}\\ y^{-7/5}\left(\dfrac{-x+2y+2}{2x+2y-1}\right)^{3/5} \end{pmatrix} =\begin{pmatrix}\dfrac{-2x+4y+1}{4x+y-2}\\ \dfrac{x-2y+4}{4x+y-2} \end{pmatrix}. \qquad(4)$$ To use $(4),$ there is more convenient returning to the source unknowns set.

$\color{green}{\textbf{Returning to the source unknowns set.}}$

Using $(2),$ we can obtain $$\frac{4x+y-2}3 = \frac{4(3-a)+3-b-2(3-c)}{3(3-c)} = \frac{9-4a-b+2c}{3(3-c)} = \frac{9-(a+b+c)+3c-3a}{3(3-c)} = \frac{c-a+2}{3-c},$$ $$\frac{x-2y+4}3 = \frac{(3-a)-2(3-b)+4(3-c)}{3(3-c)} = \frac{9-a+2b-4c}{3(3-c)} = \frac{9-(a+b+c)+3b-3c}{3(3-c)} = \frac{b-c+2}{3-c},$$ $$\frac{2x+2y-1}3 = \frac{2(3-a)+2(3-b)-(3-c)}{3(3-c)} = \frac{9-2a-2b+c}{3(3-c)} = \frac{9-(a+b+c)+3a-3b}{3(3-c)} = \frac{c-a+2}{3-c}.$$ Taking in account $(2)-(4),$ we can obtain $$\begin{cases} \left(\dfrac{a+1}{3-c}\right)^{3/5}(3-c)^2(c-a+2) = t\\[4pt] \left(\dfrac{b+1}{3-a}\right)^{3/5}(3-a)^2(a-b+2) = t\\[4pt] \left(\dfrac{c+1}{3-b}\right)^{3/5}(3-b)^2(b-c+2) = t, \end{cases}$$ where $t$ is some constant.

The final optimization can use this as condition for stationary points.

$\color{green}{\textbf{Final optimization.}}$

For $$\Phi(a,b,c)=\sum_\bigcirc\left(\frac{a+1}{3-c}\right)^{2/5}$$ $$t\Phi(a,b,c) = \sum_\bigcirc\,(a+1)(3-c)(c-a+2) = \sum_\bigcirc (a^2(c-b-4)+2ab+4a) + 18,$$ $$t\Phi'_a(a,b,c) = -2 a (b - c + 4) + b^2 + 2 (b + c) - c^2 + 4 = 0,$$ $$t\sum_\bigcirc \Phi'_a(a,b,c) = 12 - 4(a+b+c) = 0,$$ $$t\Phi'_a(a,b,c) = -2 a (b - c + 4) + (3-a)(b-c+2) + 4 = (a-1)(3c-3b-10)=0.$$ Since $3c<3b+10,$ then, taking in account the symmetry, the single stationary point within the area is $(1,1,1)$.

Thus, $$\color{green}{\mathbf{\boxed{\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25}\geq3.}}}$$

Answer 3

Let $\frac{a+1}{a+b}=x^5,$ $\frac{b+1}{b+c}=y^5$ and $\frac{c+1}{c+a}=z^5$.

Thus,$\frac{4a+b+c}{a+b}=3x^5,$ $\frac{4b+c+a}{b+c}=3y^5$ and $\frac{4c+a+b}{c+a}=3z^5$, which says that the system $$\left\{\begin{array}{ccc}(3x^5-4)a+(3x^5-1)b-c=0\\-a+(3y^5-4)b+(3y^5-1)c=0\\(3z^5-1)a-b+(3z^5-4)c=0\end{array}\right.$$ has infinitely many solutions $(a,b,c)$, which says $\Delta=0$ or $$\prod_{cyc}(3x^5-4)+\prod_{cyc}(3x^5-1)-1+\sum_{cyc}(3x^5-4)(3y^5-1)=0$$ or $$3x^5y^5z^5+2\sum_{cyc}(x^5-x^5y^5)-3=0$$ and we need to prove that $$x^2+y^2+z^2\geq3,$$ which is wrong at the general.

But as River Li said, it's enough to prove the last inequality for the condition and for $x^5y^5z^5\geq\frac{25}{27}.$

Firstly, we'll prove that $x^5y^5z^5\geq\frac{25}{27}.$

Indeed, we need to prove that $$\prod_{cyc}\frac{a+1}{a+b}\geq\frac{25}{27}$$ or $$\prod_{cyc}\frac{4a+b+c}{a+b}\geq25$$ or $$\sum_{cyc}\left(a^3-a^2b-a^2c+\frac{7}{3}abc\right)\geq0,$$ which is true by Schur.

Now, let $f(x,y,z)=x^2+y^2+z^2-3+\lambda\left(3x^5y^5z^5+2\sum\limits_{cyc}(x^5-x^5y^5)-3\right).$

Thus, in the minimum point $(x,y,z)$ we obtain $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0.$$ Thus, $$2x+5\lambda x^4(3y^5z^5+2-2y^5-2z^5)=0$$ or $$2x^2+5\lambda x^5(3y^5z^5+2-2y^5-2z^5)=0.$$ Now, let $x\neq y$ and $x\neq z$ and since $$2y^2+5\lambda y^5(3x^5z^5+2-2x^5-2z^5)=0,$$ we obtain: $$2(x+y)+10\lambda(x^4+x^3y+x^2y^2+xy^3+y^4)(1-z^5)=0.$$ By the same way $$2(x+z)+10\lambda(x^4+x^3z+x^2z^2+xz^3+z^4)(1-y^5)=0,$$ which gives $$\frac{(x^4+x^3y+x^2y^2+xy^3+y^4)(1-z^5)}{x+y}=\frac{(x^4+x^3z+x^2z^2+xz^3+z^4)(1-y^5)}{x+z}$$ or $$(y-z)\sum_{sym}\left(x^5y^4+x^5y^3z+\frac{1}{2}x^5y^2z^2+x^4y^4z+2x^4y^3z^2+\frac{1}{3}x^2y^2z^2+x^3y+\frac{1}{2}x^2y^2+x^2yz\right)=0,$$ which gives $y=z$.

Id est, it's enough to prove that $x^2+y^2+z^2\geq3$ for equality case of two variables.

Now, for $z=y$ we obtain: $$3x^5y^{10}-2\left(2x^5y^5+y^{10}\right)+2(x^5+2y^5)=3$$ or $$x^5=\frac{2y^{10}-4y^5+3}{3y^{10}-4y^5+2},$$ which gives $$\frac{2y^{10}-4y^5+3}{3y^{10}-4y^5+2}\cdot y^{10}>\frac{25}{27}$$ or $$y>\sqrt[5]{\frac{3-\sqrt{13}+2\sqrt{3+\sqrt{13}}}{6}}=0.945539...$$ and it remains to prove $$\sqrt[5]{\left(\frac{2y^{10}-4y^5+3}{3y^{10}-4y^5+2}\right)^2}+2y^2\geq3$$ and since the last inequality is true for any $y>0.9436$, we are done.

Answer 4

Partial Hint :

With your work one can show that :

$$\left(\frac{a+1}{3-x-a+a} \right)\left(\frac{3-x-a+1}{3-x-a+x}\right)\left(\frac{x+1}{x+a} \right) \geqslant 1 $$

With $1\leq a\leq 2$ and $x\in[2-a,1]$

I cannot prove it but using Bernoulli's inequality we have :

$$ \left(\frac{a+1}{x+a} \right)^{\frac25}+\frac{1}{1+\frac{2}{5}\left(\frac{3-x-a+a}{3-x-a+1}-1\right)}+\left(\frac{x+1}{3-x-a+x} \right)^{\frac25}\geq 3$$

With $x\in[0,1]$ and $1\leq a\leq 1.5$