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Algebra Level pending

If $a$ and $b$ are positive real numbers and $a+b=1$ , then the minimum of $a^4+b^4$ is $\dfrac xy$ , where $x$ and $y$ are coprime positive integers. Find the value of $x+y$ .

The answer is 9.

1 solution

Brian Charlesworth
Mar 16, 2017

Using Jensen's Inequality with $f(z) = z^{4}$ , we have that

$\dfrac{a^{4} + b^{4}}{2} \ge \left(\dfrac{a + b}{2}\right)^{4} \Longrightarrow a^{4} + b^{4} \ge 2 \times \left(\dfrac{1}{2}\right)^{4} = \dfrac{1}{8}$ .

The minimum is achieved when $a = b = \dfrac{1}{2}$ , so $x + y = 1 + 8 = \boxed{9}$ .

The next to next to next power

The answer is 9.

1 solution

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