= Solution
For $p\geq2$, integrate the stated scalar inequality to obtain the <Clarkson inequality>
$$
\left\lVert\frac{f+g}{2}\right\rVert_p^p
+\left\lVert\frac{f-g}{2}\right\rVert_p^p
\leq\frac{\lVert f\rVert_p^p+\lVert g\rVert_p^p}{2}.
$$
If $f,g$ belong to the unit ball and $\lVert f-g\rVert_p\geq\varepsilon$, then
$$
\left\lVert\frac{f+g}{2}\right\rVert_p
\leq\left(1-(\varepsilon/2)^p\right)^{1/p}<1.
$$
Thus $L^p[0,1]$ is <uniformly convex Banach space>[uniformly convex].
Now let $X$ be uniformly convex. It is enough to show that every $\Phi\in S_{X^{**}}$ lies in the canonical image of $X$. Given $\varepsilon>0$, choose the corresponding uniform-convexity constant $\delta$, and choose $f\in B_{X^*}$ with $\Phi(f)>1-\delta$. If $x,y\in B_X$ both satisfy $f(x),f(y)>1-\delta$, then
$$
\left\lVert\frac{x+y}{2}\right\rVert>1-\delta,
$$
so $\lVert x-y\rVert<\varepsilon$. By <Goldstine theorem>, every weak-star neighbourhood of $\Phi$ contains some $Jx$ with $x\in B_X$. Directing these neighbourhoods produces a norm-Cauchy net $(x_\alpha)$; completeness gives $x_\alpha\to x\in B_X$, and weak-star convergence then gives $Jx=\Phi$. Scaling handles the whole bidual ball, so $X$ is <reflexive Banach space>[reflexive].
For $p\geq2$, uniform convexity therefore makes $L^p$ reflexive. If $1<p<2$, its conjugate exponent $q$ is greater than two, so $L^q$ is reflexive. Since $(L^p)^*=L^q$ and a Banach space whose dual is reflexive is itself reflexive, $L^p$ is reflexive for every $1<p<\infty$.
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