For continuous $g$, prove that $1+\left(\int^{1}_{0}g^2(x)dx\right)^3>\left(\int^{1}_{0}g(x)\right)^3$ 
If $g:[0,1]\rightarrow \mathbb{R}$ be a continuous function Then 
  Prove that $$1+\bigg(\int^{1}_{0}g^2(x)dx\bigg)^3>\bigg(\int^{1}_{0}g(x)dx\bigg)^3$$

My attempt:
$\bullet\;\; $If $|g(x)|\geq 1,$ Then $\displaystyle \int^{1}_{0}g^2(x)dx\geq \int^{1}_{0}g(x)dx$
$$\bigg(\int^{1}_{0}g^2(x)dx\bigg)^3\geq \bigg(\int^{1}_{0}g(x)dx\bigg)^3$$
$$1+\bigg(\int^{1}_{0}g^2(x)dx\bigg)^3> \bigg(\int^{1}_{0}g(x)dx\bigg)^3$$
$\bullet\; $ If $|g(x)|<1,$ Then $\displaystyle 1>\bigg(\int^{1}_{0}g(x)dx\bigg)^3$
$$1+\bigg(\int^{1}_{0}g^2(x)dx\bigg)^3>\bigg(\int^{1}_{0}g(x)dx\bigg)^3$$
Thus, I've proven the claim. Please comment on my solution.
If the above is wrong, then how can I fix it?
 A: Your proof is wrong because you may have $|g(x)| \geq 1$ for some values of $x$ and $|g(x)| < 1$ for some other values of $x$.
Here is  a valid proof: By Holder's/ Cauchy-Schwarz inequality $(\int_0^{1} g(x) dx)^{2} \leq \int_0^{1} g^{2}(x)dx$. So it remains to see that $y^{3/2} <1+y^{3}$ where $y =\int_0^{1} g^{2}(x)dx$. Now consider the function $f(y)=1+y^{3} -y^{3/2}$. Check that this function attains its minimum of $[0, \infty)$ at the point $y=\frac  1{2^{2/3}}$ and that the value at this point is $>0$. Hence $f(y) >0$ for all $y$.  That finishes the proof. 
A: The problem: When you take each case, say, $|g(x)| \le 1$, you're still doing $\int_{[0,1]} g(x) dx$ and not something like $\int_{[0,1] \cap A} g(x) dx$ where $A = \{x \in [0,1] | |g(x)| \le 1\}$. What you've done is show the inequality is true first for the case when the image of $g$ is a subset $(-\infty,-1] \cup [1,\infty)$ and second for the case when the image of $g$ is a subset of $(-1,1)$. But these two cases are not the only possible cases. The image of $g$ could be like $[-7,\frac12]$ (I believe the image has to be a closed interval), a subset of neither $(-\infty,-1] \cup [1,\infty)$ nor $(-1,1)$. Try $g$ as the line from the point (not interval!) $(0,-7)$ to the point (not interval!) $(1,\frac12)$.
What I think it sounds like you want to try to do: Decompose $g$ into $g = g1_A + g1_{A^c}$ where $1_B: [0,1] \to \mathbb R$ is indicator function on the set $B$.


*

*For example, $|x| = |x|1_{(-\infty,0)} + |x|1_{[0,\infty)}$ $= (-x)1_{(-\infty,0)} + (x)1_{[0,\infty)}$ because $|x|1_{(-\infty,0)}$ $= (-x)1_{(-\infty,0)}$ and $|x|1_{[0,\infty)}$ $= (x)1_{[0,\infty)}$

*Maybe this could work, but I think it's a big headache. For example, try expanding $(g1_A + g1_{A^c})^3$.
Typically: The way to go about problems like these is to use an inequality that relates the square of an integral $(\int_a^b g(x) dx)^2$ to the integral of a square $\int_a^b g^2(x) dx$ or more generally the power of an integral $(\int_a^b g(x) dx)^p$ to the integral of a power $\int_a^b g^p(x) dx$. I believe the most common are Minkowski's inequality, Hölder's inequality and the Cauchy–Bunyakovsky–Schwarz inequality.
Let's try the following version of Hölder's inequality:

Let $a,b \in \mathbb R$ with $a < b$. Let $f,g: [a,b] \to \mathbb R$
  be functions that are continuous over their common domain $[a,b]$. Let
  $p,q \in [1,\infty)$ with $\frac1p + \frac1q = 1$
$$\int_a^b |fg| dx \le \left(\int_a^b |f|^p dx\right)^{\frac1p} \left(\int_a^b |g|^q dx\right)^{\frac1q}$$

By choosing $f$ as the constant function with value $1$ and choosing $p=q=2$, we get:
$$\int_a^b |g| dx \le \left(b-a\right)^{\frac12} \left(\int_a^b |g|^2 dx\right)^{\frac12} = \left(b-a\right)^{\frac12} \left(\int_a^b g^2 dx\right)^{\frac12}$$
Assuming $b-a=1$, we get:
$$\int_a^b |g| dx \le \left(\int_a^b g^2 dx\right)^{\frac12} \tag{A}$$
$\int_a^b |g| dx \ge 0$ since $|g|(x):=|g(x)| \ge 0$ for all $x \in [a,b]$. Then $(A)$ is equivalent to $(B)$ as follows:
$$\left(\int_a^b |g| dx\right)^2 \le \int_a^b g^2 dx \tag{B}$$
Anyway, I think I won't really use $(B)$. Raising both sides of $A$ to $3$, we get:
$$\left(\int_a^b |g| dx \right)^3 \le \left(\int_a^b g^2 dx\right)^{\frac32} \tag{C}$$
Since $g(x) \le |g|(x):=|g(x)|$ for all $x \in [a,b]$, $\int_a^b g dx \le \int_a^b |g| dx $ and then:
$$\left(\int_a^b g dx \right)^3 \le \left(\int_a^b |g| dx \right)^3 \tag{D}$$
Combining $(C)$ and $(D)$, we get:
$$\left(\int_a^b g dx \right)^3 \le \left(\int_a^b g^2 dx\right)^{\frac32} \tag{E}$$
If we somehow show $(F)$ as follows, then combining $(E)$ and $(F)$ and letting $a=0$ and $b=1$ yields the desired inequality.
$$\left(\int_a^b g^2 dx\right)^{\frac32} \le 1 + \left(\int_a^b g^2 dx\right)^3 \tag{F}$$
To show $(F)$: Let $c \in \mathbb R$. The inequality $c^{\frac 32} \le 1+c^3$ is true if (and only if I guess) $c \ge 0$. Here, we have $c=\int_a^b g^2 dx \ge 0$ since $(g^2)(x):=(g(x))^2 \ge 0$ for all $x \in [a,b]$.
Remark: The desired inequality is true for $0$ and $1$ replaced with, respectively, $a$ and $b$ for any real numbers $a$ and $b$ where $b-a=1$
How to prove $c^{\frac 32} \le 1+c^3$ if $c \ge 0$ without wolfram alpha: Consider the function $f: [0,\infty) \to \mathbb R$, $f(c) := 1+c^3 - c^{\frac 32}$. We want to show that $image(f)=f([0,\infty)) \subseteq [0,\infty)$, i.e. $f(c) \ge 0$ for all $c \ge 0$. Observe that $f$ is second-differentiable. Then use first and second derivatives to show that $f$ has a global minimum at some point $(c_0,f(c_0))$. If $f(c_0) \ge 0$ (Note: Actually, $f(c_0)$ isn't zero, but even if it were, then that would be okay since we want to show $f \ge 0$ not $f > 0$), then $f(c) \ge f(c_0) \ge 0$ for all $c \in Domain(f) = [0,\infty)$
