# Complicated Constraint - Simple Inequality

### Solution 1

We know that

\displaystyle\begin{align} (a+b+c)^3 &= \sum_{cycl}a^3+\prod_{cycl}(a+b)\\ &\ge \sum_{cycl}a^3+\frac{8}{\sqrt[8]{a^3+b^3+c^3}}\\ &\ge (1+8)\sqrt[9]{ \sum_{cycl}a^3\left( \frac{1}{\sqrt[8]{a^3+b^3+c^3}}\right)^8}\\ &=9. \end{align}

Hence $a+b+c\ge 9^{1/3}.$

The equality is attained only if $a=b=c=\displaystyle\frac{1}{\sqrt[3]{3}}.$

### Solution 2

The given condition can be rewrtitten as

(1)

$\displaystyle\sqrt[8]{(a^3+b^3+c^3)(a+b)^8(b+c)^8(c+a)^8\left(\frac{3}{8}\right)^8}\ge 1.$

Applying the AM-GM inequalty, we obtain

\displaystyle\begin{align} &\sqrt[8]{(a^3+b^3+c^3)\underbrace{\frac{3}{8}(a+b)(b+c)(c+a)\cdots\frac{3}{8}(a+b)(b+c)(c+a)}_{8\,\text{times}}}\\ &\le\frac{1}{9}\left[(a^3+b^3+c^3)+\underbrace{\frac{3}{8}(a+b)(b+c)(c+a)+\ldots+\frac{3}{8}(a+b)(b+c)(c+a)}_{8\,\text{times}}\right]\\ &=\frac{1}{9}\left[(a^3+b^3+c^3)+3(a+b)(b+c)(c+a)\right]\\ &=\frac{1}{9}(a+b+c)^3. \end{align}

Combining this with (1) we infer that

$(a+b+c)^3\ge 9,$

and the result follows.

The equality occurs if and only if $a=b=c=\displaystyle\frac{1}{\sqrt[3]{3}}.$

### Acknowledgment

The problem above (from the Romanian Mathematical Magazine, #SP036, posed December 14, 2016) has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru, along with two solutions. The problem has been proposed by Nguyen Viet Hung. Two practically identical solutions (Solution 1) have been submitted by Anas Adlany (Morocco) and Soumitra Mandal (India). Solution 2 is by Nguyen Viet Hung. The illustration is by Gary Davis.