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Algebra Difficulty 7.6 National Olympiad, round 2 Prove it Slovenia

Prove that for any positive real numbers $a$ , $b$ and $c$ the following holds:
$a + \sqrt{ab} + \sqrt[3]{abc} \le \frac{4}{3}(a + b + c).$

Solution

Recall that the inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list of numbers. From this we get
$a + \sqrt{ab} + \sqrt[3]{abc} = a + \sqrt{\frac{a}{2} \cdot 2b} + \sqrt[3]{\frac{a}{4} \cdot b \cdot 4c} \le a + \frac{1}{2}\left(\frac{a}{2} + 2b\right) + \frac{1}{3}\left(\frac{a}{4} + b + 4c\right) = \frac{4}{3}(a+b+c).$
Here, equality holds if and only if $a = 4c = 16c$ .

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