At least one of $|f(x)|$ and $|g(x)|$ not less than $a+1$ Let $a\in(0,1),f(x)=ax^3+(1-4a)x^2+(5a-1)x-5a+3$,$g(x)=(1-a)x^3-x^2+(2-a)x-3a-1 $.
Prove that:

For any real number $x$ ,at least one of $|f(x)|$  and $|g(x)|$ not less than $a+1$

since
$$f(x)+g(x)=x^3-4ax^2+(1+4a)x-8a+2$$
if we can prove 
$$|f(x)+g(x)|\ge 2(a+1)$$?but can not find any other and can not prove whether there is any other solution or not.
HINT: $max\{f,g\}=\frac{f+g+|f-g|}{2}$
and the problem is wrong. counterexample:
$f(1)=3-3a\geq a+1 \Rightarrow a \leq {1\over2}$ and
$g(1)=1-5a\geq a+1 \Rightarrow a \leq 0$ so if $a&gt;1/2$ both f(1) and g(1) less than a+1
Suppose $|f|, |g| &lt; a+1$ on $\mathbb{R}$. Then
$$
(f+g)(x) = x^{3} - 4ax^{2} + (1+4a)x - 8a + 2 &lt; 2(a+1) &lt; 4
$$
for all $x \in \mathbb{R}$.
But
$(f+g)(x) \sim x^{3}$ as $x \to \infty$,
$\to \gets$.