Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ and $a+b+c=1$.

Prove that: $$\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\leq\frac{31}{27}$$

The equality occurs for $(a,b,c)=\left(\frac{19}{27},\frac{8}{27},0\right)$.

This inequality is similar to the following inequality, which was proposed by Walther Janous.

For all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\neq0$ prove that: $$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq\frac{5}{4}\sqrt{x+y+z}$$

My proof:

By Cauchy-Schwarz $$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$ Id est, it remains to prove that $$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or $$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$ $$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or $$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$ $$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.

If we want to use a similar way for the starting inequality, we need to use Holder, which gives very big numbers.

Maybe there is another reasoning?

Thank you!

  • $\begingroup$ Since $c=1-a-b$ how about rewriting the LHS of the inequality as a function $f(a,b)$ which defines a surface. Then find maxima of the surface. $\endgroup$
    – Χpẘ
    Apr 9, 2017 at 21:51
  • $\begingroup$ So, you have a Holder solution although the coefficients are very big? $\endgroup$
    – River Li
    Jul 3, 2019 at 4:23
  • $\begingroup$ @River Li No, I have no solution. I can get inequality after using Holder, but I can not prove it. $\endgroup$ Jul 3, 2019 at 4:26

3 Answers 3


With computer, here is an ugly solution:

WLOG, assume that $c = \min(a, b, c).$ Since $x^{2/3}, x \ge 0$ is concave, we have \begin{align} \frac{b}{\sqrt[3]{b+c}} + \frac{c}{\sqrt[3]{c+a}} &= \frac{b}{b+c}(b+c)^{2/3} + \frac{c}{c+a}(c+a)^{2/3}\\ &\le \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^{2/3}. \end{align} On the other hand, we have $$\Big(\frac{1+5a+5b}{5+a+b}\Big)^3 - (a+b)^2 = \frac{(a^2+2ab+b^2+18a+18b+1)(1-a-b)^3}{(5+a+b)^3} \ge 0$$ which results in $(a+b)^{2/3} \le \frac{1+5a+5b}{5+a+b}$.

Thus, it suffices to prove that $$\frac{a}{a+b}\frac{1+5a+5b}{5+a+b} + \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^{2/3} \le \frac{31}{27}.$$ It suffices to prove that $$\Big(\frac{31}{27} - \frac{a}{a+b}\frac{1+5a+5b}{5+a+b}\Big)^3\ge \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)^3\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^2. $$ After homogenization, it suffices to prove that $$(a+b+c)^2\Big(\frac{31}{27} - \frac{a}{a+b}\frac{(a+b+c)+5a+5b}{5(a+b+c)+a+b}\Big)^3\ge \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)^3\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^2.$$ If suffice to prove that $f(a,b,c) \ge 0$ for $a, b, c \ge 0; \ c = \min(a,b,c)$ where $f(a,b,c)$ is a polynomial (a long expression).

We apply the Buffalo Way. There are three possible cases:

1) If $c = 0$, we have $f(a,b, 0) = 216a(a+28b)(8a-19b)^2(a+b)^5 \ge 0$.

2) If $c > 0$ and $c \le a \le b$, let $a = c + s, \ b = c+s+t; \ s,t \ge 0$. $f(c+s, c+s+t, c)$ is a polynomial in $c, s, t$ with non-negative coefficients.

3) If $c>0$ and $c \le b\le a$, let $b = c+s, \ a = c+s + t; \ s,t \ge 0$. We have \begin{align} &f(c+s+t, c+s, c) = g(c,s,t) + 216(s+t)(29s+t)(11s-8t)^2(2s+t)^5\\ &\quad + 108(146777s^4+13368s^3t-97257s^2t^2+42286st^3+7200t^4)(2s+t)^4c \end{align} where $g(c,s,t)$ is a polynomial with non-negative coefficients. Note that $$146777s^4+13368s^3t-97257s^2t^2+42286st^3+7200t^4 \ge 2\sqrt{146777 \cdot 7200} s^2 t^2 + 2\sqrt{13368\cdot 42286}s^2t^2 - 97257s^2t^2 \ge 0.$$

The desired result follows. We are done.

  • $\begingroup$ Your solution is the most beautiful one here.. $\endgroup$
    – user822157
    Feb 19, 2021 at 8:38
  • 2
    $\begingroup$ @haidangel Thanks. The BW is not nice. $\endgroup$
    – River Li
    Feb 19, 2021 at 9:10

Here is my personal take on the problem;

We can use the substitutions $b+c = 1-a$, $c+a = 1-b$, $c = 1-a-b$ to reduce the problem to the following inequality in arbitrary $a,b$;

$$\dfrac{a}{\sqrt[3]{a+b}} + \dfrac{b}{\sqrt[3]{1-a}} + \dfrac{1-a-b}{\sqrt[3]{1-b}} \leq \dfrac{31}{27}$$

Define $f(a,b) = \dfrac{a}{\sqrt[3]{a+b}} + \dfrac{b}{\sqrt[3]{1-a}} + \dfrac{1-a-b}{\sqrt[3]{1-b}}$. I suspect that using the first and second derivative tests to find local maxima-minima could solve the problem from here, as $\dfrac{\partial f}{\partial a}$ and $\dfrac{\partial f}{\partial b}$ are, while messy, not difficult to calculate.

In fact, this can be generalized quite a bit. This method can be generalized as follows;

Consider $\sum_{cyc} g(x_1,\dots,x_n)$ for some well-defined function $g$, where the cyclic behavior occurs over a set of variables $x_1,\dots,x_k$ for $k > n$ and we restrict these $x$ by stating that $\sum_{j=1}^k x_j = A$ for a fixed constant $A$. Then we can write $x_k = A - \sum_{j=1}^{k-1} x_j$ and use this as the basis for a substitution to create a function $f(x_1,\dots,x_{k-1}) = \sum_{cyc} g(x_1,\dots,x_n)$ with restrictions on the $x_j$ removed and using the substitution previously given for $x_k$. If $g$ is differentiable, then so will $f$ be, and so local extremes can be calculated. If $g$ is twice differentiable, then so is $f$ and we can use the second derivative test to test whether these values are true extrema or saddle points.


I have an half proof :

We have :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have : $$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$

We need a lemma :


Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have : $$\sum_{cyc}0.5(a+b)^{\frac{2}{3}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$

Proof : It's Jensen's inequality apply to the concave function $x^{\frac{2}{3}}$.

Now we prove :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have : $$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \frac{a+b}{2(a+b)^{\frac{1}{3}}}+\frac{b+c}{2(b+c)^{\frac{1}{3}}}+\frac{a+c}{2(a+c)^{\frac{1}{3}}}$$

Proof : The inequality is equivalent to :

$$\frac{b-a}{2(a+b)^{\frac{1}{3}}}+\frac{c-b}{2(c+b)^{\frac{1}{3}}}+\frac{a-c}{2(a+c)^{\frac{1}{3}}}\leq 0$$

Wich is true with SOS method or rearrangement inequality .

Finally we have :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have : $$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}} <\frac{31}{27}$$

Conclusion : So the real case to threat is when $b\geq a\geq c $.

  • 1
    $\begingroup$ Your proof is wrong. Try $(a,b,c)=\left(\frac{19}{27},\frac{8}{27},0\right).$ $\endgroup$ Oct 25, 2019 at 12:27
  • $\begingroup$ Sorry it was a problem of notation with the cyclic sum...Now it works ! $\endgroup$
    – Erik Satie
    Oct 25, 2019 at 13:00
  • 1
    $\begingroup$ @The.old.crap Actually, you proved the original inequality for $b\ge a\ge c$. $\endgroup$
    – River Li
    Oct 25, 2019 at 13:52

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