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

$a,b,c >0$ 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 $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1$$ :(

• a=1 b=1 c=1 equations is true Commented May 20, 2016 at 20:42
• What was your counter example? Also, maybe $a+b=3-c$ will help. Commented May 20, 2016 at 20:45
• $a=1.1, b=1.1,c =0.8$ Commented May 20, 2016 at 20:51
• Hard to tell from the first look, but I think the problem is that you do not use the fact that the exponent is $\frac{2}{5}$ in your proof. So, using your method we can 'prove' the first inequality for any exponent, which can't be true Commented Jun 2, 2016 at 12:46
• @HN_NH The Buffalo Way (BW) works. But I am afraid it is not what you want. Commented Nov 27, 2020 at 1:59

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

Assuming $$0, 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}

Now assuming $$0. 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 $$75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge0$$.

– user822157
Commented Mar 1, 2021 at 4:39

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

For $$c=0,\ b=3-a$$ and WLOG $$0 we have the inequality $$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, $$\color{green}{\mathbf{\boxed{\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25}\geq3.}}}$$

• You really like this kind of inequalities don't you :) Commented Sep 5, 2016 at 19:21
• but these problems are for high school students only. The use of Lagrange multiplier or derivative with root finder from computer, or numerical methods are cheap and defeat the purpose of the problem. Commented Sep 5, 2016 at 21:34
• @YuriNegometyanov, this would sound harsh, but before you lecture other people about "success", maybe you should have checked your algebra 101 before posting these mechanical proofs, e.g., (1) when $c=0$, $$\frac{b+1}{b+c}=\frac{\mathbf{4}-a}{3-a}$$ (2) when $A+B+C=3$, $$\frac{2-A}{C}=\frac{2(A+B+C)-\mathbf{3}A}{3C}$$ To whoever up-voted, have you actually read the "proof"?? Commented Sep 12, 2016 at 15:41
• I think it's not answer. Commented Nov 5, 2017 at 18:33
• @YuriNegometyanov, +1 ; does \sum\bigcirc mean $\sum\limits_{cyc},$ thank you so much.
– user822157
Commented Mar 1, 2021 at 4:38

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.

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 $$1\leq a\leq 1.5$$

• @RiverLi Any idea to finish it ? Can you correct if my reasoning is wrong ? Thanks a lot !! Commented Dec 2, 2020 at 13:16
• You only give partial result with some additional assumption. By the way, you should write your proof clearly. For example, you wrote "I forgot to mention that to works that we need ...", this is bad, you should write it in the beginning. Commented Dec 2, 2020 at 15:09
• @RiverLi I give up it's too hard ! Without BW I don't see any progress... Commented Dec 2, 2020 at 19:13
• @RiverLi a last question have you a counter-example for the inequality $f(x)\geq 6-g(x)$ ? Commented Dec 2, 2020 at 20:02
• Do you mean $f(x) = A^{2/5} + B^{2/5} + C^{2/5}$ and $g = \dfrac{1}{1 + \frac{2}{5}(\frac{1}{A}-1)} + \cdots$. Try $a = 1.1$ and $x=0$. Commented Dec 3, 2020 at 1:46