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 
 $$\frac{1}{3}e^{2t} + \frac{2}{3}e^{-t}\leq e^{2t^2}?$$
 A: For $t\geq1$
$$\dfrac{1}{3}e^{2t}+\dfrac{2}{3}e^{-t}\leq\dfrac{1}{3}e^{2t}+\dfrac{2}{3}e^{2t}=e^{2t}\leq e^{2t^2}$$
For $t\leq-1$
$$\dfrac{1}{3}e^{2t}+\dfrac{2}{3}e^{-t}\leq\dfrac{1}{3}e^{-t}+\dfrac{2}{3}e^{-t}=e^{-t}\leq e^{2t^2}$$
I need to think of something for $-1<t<1$
A: For $|t|\gt1$, ${1\over3}e^{2t}+{2\over3}e^{-t}\lt{1\over3}e^{2t^2}+{2\over3}e^{2t^2}=e^{2t^2}$ is obvious, so we need only worry about $|t|\le1$.  Combining
$$e^{2t}=1+2t+2t^2+{4\over3}t^3+{2\over3}t^4+{4\over15}t^5+\cdots$$
and 
$$e^{-t}=1-t+{1\over2}t^2-{1\over6}t^3+{1\over24}t^4-{1\over120}t^5+\cdots$$
gives
$$\begin{align}
{1\over3}e^{2t}+{2\over3}e^{-t}
&=1+t^2+{1\over3}t^3+{1\over4}t^4+{1\over12}t^5+\cdots\\
&=1+t^2+t^2\left({1\over3}t+{1\over4}t^2+{1\over12}t^3+\cdots\right)\\
&\le1+t^2+t^2\left({1\over2}+{1\over4}+{1\over8}+\cdots\right)\\
&=1+2t^2\\
&\le e^{2t^2}
\end{align}$$
The main simplifying inequality here is
$${1\over3}\left({2^n\pm2\over n!} \right)\le{1\over2^{n-2}}$$
for $n\ge3$.
A: We have to prove that:
$$ 2t^2 \geq \log\left(e^{3t}+2\right)-\log(3e^t) $$
hence, by switching to derivatives, it is enough to prove that
$$ \frac{e^{3 t} - 1}{2+e^{3 t}}\leq 2t $$
holds for every $t\geq 0$ (and the opposite happens for $t<0$). That is trivial for $t\geq\frac{1}{2}$. 
Since $e^{3t}\geq 1+3t$, it is enough to prove that over $\left[0,\frac{1}{2}\right]$ we have:
$$ e^{3t}\leq 1+6t+6t^2, $$
but that follows from a simple convexity argument:
$$ \frac{e^{3t}-1}{3t}\leq 1+2t\left(\frac{e^{3/2}-1}{3/2}-1\right)\leq 1+3t\leq  2+2t. $$
A: 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 $t\neq 0$.
A: For $0<t<1$, it's obvious that $$\dfrac{2}{3}\leq\dfrac{2}{3}e^{t}$$ and $$e^{3t} < e^{2t^2+t}$$.With these two facts, we may immediately get $$\dfrac{1}{3}e^{3t}+\dfrac{2}{3}\leq e^{2t^2+t}.$$ Then, multiply both sides by $$ e^{-t}$$.
A: Don't prove it for every $t$. Prove it for every $\log t$. It just so happens that every real number is the $\log$ of some real number, so this is sufficient...
