= Solution
A real <random variable> is a <sub-Gaussian random variable> with exponent $b\geq0$ when its <moment-generating function> satisfies
$$
\mathbb E e^{tX}\leq e^{b^2t^2/2}\qquad(t\in\mathbb R).
$$
Differentiating at $t=0$ from both sides gives $\mathbb EX=0$. Exponential integrability near zero justifies this differentiation. Conversely, if $|X|\leq M$ and $\mathbb EX=0$, <convexity> gives
$$
e^{tX}\leq\frac{M+X}{2M}e^{tM}+\frac{M-X}{2M}e^{-tM},\qquad
\mathbb Ee^{tX}\leq\cosh(tM)\leq e^{t^2M^2/2}.
$$
The last inequality follows by comparing power series using $(2j)!\geq2^j j!$. The case $M=0$ is immediate. Thus \b[a bounded <random variable> is sub-Gaussian exactly when it is centered]; the bound $M$ is one possible exponent.
For $b>0$, the <Markov inequality> applied to $e^{tX}$ gives $\mathbb P(X>R)\leq\exp(b^2t^2/2-tR)$. Choose $t=R/b^2$ and repeat with $-X$. The <union bound> yields
$$
\boxed{\mathbb P(|X|>R)\leq2e^{-R^2/(2b^2)}}\qquad(R>0).
$$
If $b=0$, the same exponential estimate with arbitrarily large $t$ shows $X=0$ almost surely, so all subsequent inequalities are trivial.
Integrating this <tail probability> estimate gives, for every real $k\geq2$,
$$
\mathbb E|X|^{2k}=2k\int_0^\infty r^{2k-1}\mathbb P(|X|>r)\,dr
\leq2(2b^2)^k\Gamma(k+1).
$$
We need $2\Gamma(k+1)\leq k^k$. For integer $k$ this follows inductively from $2\cdot2!=2^2$ and $(k+1)k^k\leq(k+1)^{k+1}$. To cover real $k$ as well, set $H(k)=k\log k-\log\Gamma(k+1)-\log2$. Differentiating the <Gamma function> integral gives $(\log\Gamma)'(k+1)=\mathbb E\log Y$ for a <Gamma distribution> with shape $k+1$ and rate $1$. The <Jensen inequality> bounds this by $\log(k+1)$. Hence $H'(k)\geq1-\log(1+1/k)>0$ and $H(2)=0$. Taking the $2k$th root now proves
$$
\boxed{\|X\|_{2k}\leq b\sqrt{2k}\quad(k\geq2)}.
$$
Here $\|X\|_p=(\mathbb E|X|^p)^{1/p}$ is the <Lp norm>.
For the final assertion we prove the required <Littlewood interpolation inequality>, rather than assuming it. Apply <Holder inequality> with conjugate exponents $3/2$ and $3$:
$$
\mathbb E|X|^2=\mathbb E\bigl(|X|^{2/3}|X|^{4/3}\bigr)
\leq(\mathbb E|X|)^{2/3}(\mathbb E|X|^4)^{1/3}.
$$
Raising to the power $3/2$ gives $\|X\|_2^3\leq\|X\|_1\|X\|_4^2$. The preceding <Lp norm> estimate with $k=2$ says $\|X\|_4\leq2b$. Therefore
$$
\boxed{\|X\|_2^3\leq4b^2\|X\|_1}.
$$
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