How to prove that $\frac{1}{3}e^{2t} + \frac{2}{3}e^{-t}\leq e^{2t^2}$ How to prove that, for every real $t$, one has 
 $$\frac{1}{3}e^{2t} + \frac{2}{3}e^{-t}\leq e^{2t^2}?$$

Elementary but not entirely trivial: Note that for real $t$ both $t-t^2$ and $-t-2t^2$ are at most $\tfrac{1}{4}$ and therefore we can use the inequality $$e^x\leq\frac{1}{1-x}$$ that holds for all $x<1$ to obtain
$$\tfrac{1}{3}\left(e^{t-t^2}\right)^2+\tfrac{2}{3}e^{-t-2t^2}\leq \frac{1}{3(1-t+t^2)^2}+\frac{2}{3(1+t+2t^2)}.$$ The right hand side is in fact $\leq 1$, proving your original inequality.  One (tedious) way to see this is to compute
$$1-\frac{1}{3(1-t+t^2)^2}-\frac{2}{3(1+t+2t^2)}=\frac{t^2(1+(2t^2-t)(5-3t+3t^2))}{3(1-t+t^2)^2(1+t+2t^2)}$$
and show that numerator and denominator are both positive for $t\neq 0$.
