Mathematical Reflections 2017, Issue 2 - Problem J408

Problem:
Let $a$ and $b$ be nonnegative real numbers such that $a+b=1$. Prove that $$\dfrac{289}{256} \leq (1+a^4)(1+b^4) \leq 2.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
By the Cauchy-Schwarz Inequality, we have $$(1+a^4)(1+b^4) \geq (1+a^2b^2)^2=(1+a^2(1-a)^2)^2 \geq \left(1+\dfrac{1}{16}\right)^2=\dfrac{289}{256}$$ and the minimum is attained if and only if $a=b=\dfrac{1}{2}$. By the AM-GM Inequality, we have
$$(1+a^4)(1+b^4) \leq \dfrac{(1+a^4)^2+(1+b^4)^2}{2}=\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2}.$$
Using the fact $$\begin{array}{rcll} a^4 & \leq & (1-a)^4, & a \in [0,1/2] \\ (1-a)^4 & \leq & a^4, & a \in [1/2,1], \end{array}$$ then if $a \in [0,1/2]$, we have
$$\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2} \leq (1+(1-a)^4)^2 \leq 2$$
and if $a \in [1/2,1]$, we have
$$\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2} \leq (1+a^4)^2 \leq 2.$$
The conclusion follows.