Prove that $1 + e^{2 y} - e^y (2 + y^2)\geq 0$ for any $y$? It is straightforward to see that  $\displaystyle 1 + e^{2 y} - e^y (2 + y^2)>0$  for $y \in \mathbb{R}$ by plotting its graph. But is there any way to prove this mathematically?
 A: Define the function $f$ on $\mathbb{R}$ by
$$\forall x\in\mathbb{R},\ f(x)=1+\mathrm{e}^{2x}-\mathrm{e}^x\bigl(2+x^2\bigr).$$
It is easly seen that for all $x\in\mathbb{R}$,
$$f(x)=\mathrm{e}^x\Bigl(\mathrm{e}^x+\mathrm{e}^{-x}-2-x^2\Bigr)=2\mathrm{e}^x\left(\cosh(x)-1-\frac{x^2}2\right).$$
Now fix $x\in\mathbb{R}^*$.
From Taylor–Lagrange formula, we know that there exists $c$ between $0$ and $x$ such that
$$\cosh(x)=1+\frac{x^2}2+\cosh(c)\frac{x^4}{24}.$$
Hence
$$f(x)=2\mathrm{e}^x\cosh(c)\frac{x^4}{24}>0.$$
We have thus shown that:
$$\forall x\in\mathbb{R}^*,\ 1+\mathrm{e}^{2x}-\mathrm{e}^x\bigl(2+x^2\bigr)>0.$$
A: Let $f(y)=1+e^{2y}-e^y (2+y^2)$ be your function. Calculate $\partial_y f(y)=0$ to find the minimum.
$$\partial_y f(y)=2\cdot e^{2y} - 2e^{y}\cdot y - e^y (2+ y^2)=0$$
$$e^{y} - y - 1 - \frac{y^2}{2}=0$$
$$e^{y} = 1 + y + \frac{y^2}{2}$$
Now observe that Taylor series of $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)$
Therefor the right side will always be bigger than the left one, and there is no minimum for y>0. Therefore $f(y)$ the graph is always bigger than 0.
The same reasoning works for y<0. You have to check that f(0)>0, then you are done.
A: For $y\to\pm\infty$ it is obviously positive. For large $y$ the $e^{2y}$ dominates, whereas for large negative $y$ it is the $1$ which dominates. Then it is enough to ask ourselves, whether it can be equal to zero for any $y$. This results in the condition
\begin{equation}
\frac{e^y+e^{-y}}{2}=1+\frac{y^2}{2}
\end{equation}
I don't know what is the most elementary thing to do here. Both functions are positive, and obviously, equality holds for $y=0$. I would use the Taylor expansion. On the l.h.s. we have
\begin{equation}
\frac{e^y+e^{-y}}{2}=\cosh(y)=1+\frac{y^2}{2}+\frac{y^4}{4!}+\frac{y^6}{6!}+\dots
\end{equation}
For any $y\ne 0$ there are an infinite number of additional positive terms, therefore equality can only hold for $y=0$.
Without Taylor expansion I don't know what to do, but there is certainly a simple proof also.
