Solving the inequality $\sum_{cyc a, b, c}a(\sqrt[3]{b/a} -b) \leq 2/3$ with constraint $a+b+c = 1$

Question

I was doing some contest math exercises, and stumbled onto this problem

Let $a, b, c$ be positive real numbers such that $a + b+ c = 1$. Show that $$a\left(\left(\frac{b}{a}\right)^{1/3} -b\right) + b\left(\left(\frac{c}{b}\right)^{1/3} -c\right) + c\left(\left(\frac{a}{c}\right)^{1/3} -a\right) \leq 2/3$$

So far I have been unsuccessful in finding a solution. I am seeking any sort of hint to the problem, but a complete solution is also okay with me.

My observations:

1. Maybe a substitution $x, y, z = b/a, c/b, a/c$ could simplify the inequality? I don't see anything here though.

2. With a concavity argument (Jensen's inequality), one gets

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

$$\leq \sqrt[3]{a\frac{b}{a} + b\frac{c}{b} + c\frac{a}{c}} = \sqrt[3]{a+b+c} = 1$$

but I haven't been able to get anywhere with this.

Answer 1

We need to prove that: $$\sum_{cyc}a^3\left(\frac{b}{a}-b^3\right)\leq\frac{2}{3},$$ where $a$, $b$ and $c$ are positives such that $a^3+b^3+c^3=1$ or $$\sum_{cyc}a^3\sum_{cyc}a^2b-\sum_{cyc}a^3b^3\leq\frac{2}{3}(a^3+b^3+c^3)^2$$ or $$\sum_{cyc}(2a^6-3a^5b-3a^4c^2+7a^3b^3-3a^3b^2c)\geq0$$ or $$\sum_{cyc}(4a^6-6a^5b-6a^4c^2+14a^3b^3-6a^3b^2c)\geq0$$ or $$\sum_{cyc}(4a^6-3a^5b-3a^5c-3a^4b^2-3a^4c^2+14a^3b^3-3a^3b^2c-3a^3c^2b)\geq$$ $$\geq\sum_{cyc}(3a^5b-3a^5c-3a^4b^2+3a^4c^2-3a^3c^2b+3a^3b^2c)$$ or $$\sum_{sym}(2a^6-2a^5b-a^5b+a^4b^2-4a^4b^2+4a^3b^3+3a^3b^3-3a^3b^2c)\geq$$ $$\geq3\sum_{cyc}(a^5b-a^5c-a^4b^2+a^4c^2-a^3c^2b+a^3b^2c)$$ or $$\sum_{cyc}(a-b)^2(2(a^4+a^3b+a^2b^2+ab^3+b^4)-ab(a^2+ab+b^2)-4a^2b^2+3c^3(a+b))\geq$$ $$\geq3(a-b)(a-c)(b-c)\sum_{cyc}\left(a^3+a^2b+a^2c+\frac{1}{3}abc-a^2b-a^2c-\frac{2}{3}abc+\frac{1}{3}abc\right)$$ or $$\sum_{cyc}(a-b)^2(2a^4+a^3b-3a^2b^2+ab^3+2b^4+3c^3(a+b))\geq3(a-b)(a-c)(b-c)(a^3+b^3+c^3)$$ or $$\sum_{cyc}(a-b)^2(2a^4+a^3b-3a^2b^2+ab^3+2b^4+3c^3(a+b))\geq\sum_{cyc}(b-a)^3(a^3+b^3+c^3)$$ or $$\sum_{cyc}(a-b)^2(2a^4+a^3b-3a^2b^2+ab^3+2b^4+3c^3(a+b)+(a^3+b^3+c^3)(a-b))\geq0$$ or $$\sum_{cyc}(a-b)^2(3a^4-3a^2b^2+2ab^3+b^4+4c^3a+2c^3b)\geq0,$$ which is true because by AM-GM $$3a^4-3a^2b^2+2ab^3+b^4+4c^3a+2c^3b\geq a^4-3a^2b^2+2ab^3=$$ $$=a(a^3+2b^3-3ab^2)\geq a\left(3\sqrt[3]{a^3\cdot(b^3)^2}-3ab^2\right)=0.$$

Answer 2

We can rewrite the expression as $$\sum_{cyc}\sqrt[3]{a^2b}-\sum_{cyc}ab$$ By AM-GM, $$\sum_{cyc}\sqrt[3]{a^2b}\le\sum_{cyc}\frac{2a+b}3=a+b+c=1$$ We can consider $$\sum_{cyc}(a-b)^2=\sum_{cyc}a^2-\sum_{cyc}ab\ge0$$ and $$(a+b+c)^2=\sum_{cyc}a^2+2\sum_{cyc}ab=1$$ to show that $$\sum_{cyc}ab\ge\frac13$$ Therefore we can conclude that $$\sum_{cyc}a\left(\sqrt[3]{\frac{b}{a}}-b\right)\le1-\frac13=\frac23$$