Prove $h(t)=\frac{-t}{3}-e^{\frac{-t}{3}+\sqrt{2} \sqrt{t}}+t+1+\sqrt{2} \sqrt{t}>0$ when $t>0$. I need to prove that $$h(t)=\frac{-t}{3}-e^{\frac{-t}{3}+\sqrt{2} \sqrt{t}}+t+1+\sqrt{2} \sqrt{t}\ge 0$$ when $t>0$. Since $h(0)=0$, I thought I could just prove that $h'(t)\ge 0$ first. Also, the Mathematica calculation proved the assumption ($h'(t)>0$) correct. However, the derivative seems to be quite nasty and I can't find a solution. I wonder if there's a solution.
Plus, I also need to prove another inequality which seems a little bit harder.
$$g(t)=\frac{-t}{3}-e^{\frac{-t}{3}-\sqrt{2} \sqrt{t}}+t+1-\sqrt{2} \sqrt{t}>0$$ when $t>0$.

 A: Let $x=\sqrt{2t}$ and $u=x-\frac{x^2}{6}$; then you want to show that $$\frac{x^2}{2}\geq e^u-(1+u)$$ for $x\geq0$.
To proceed further, we need some bounds on $u$; by completing the square, we see that $0\leq u\leq\frac{3}{2}$ for $x\in[0,6]$, and $u\leq0$ otherwise.
When $u\leq0$, the claim is easy: $e^u\leq1$, so that $$e^u-(1+u)\leq-u=\frac{x^2}{6}-x\leq\frac{x^2}{6}\leq\frac{x^2}{2}$$
Now assume $0\leq u\leq\frac{3}{2}$.  In that case, $$e^u-(1+u)=\sum_{j=2}^{\infty}{\frac{u^j}{j!}}\leq\sum_{j=2}^{\infty}{\frac{u^j}{2!\cdot3^{j-2}}}=\frac{u^2}{2}\frac{1}{1-\frac{u}{3}}$$
Thus we want to show that $$\frac{u^2}{2}\frac{1}{1-\frac{u}{3}}\leq\frac{x^2}{2}$$  Note that $1-\frac{u}{3}>0$, since $u\leq\frac{3}{2}$.  Substituting for $u$, canceling a common factor of $\frac{x^2}{2}$, and rearranging, an equivalent claim is $$\left(1-\frac{x}{6}\right)^2\leq1-\frac{x}{3}\left(1-\frac{x}{6}\right)$$  But this is just a quadratic in $x$; collecting like terms, it suffices to show $$\frac{x^2}{36}\geq0$$ which is obvious.
A: Inspired by Manaker, another approach with derivatives can also work the problem out. Here I just prove the inequality for $g(t)$. For $h(t)$, the method is basically the same.
Let $u=-t/3-\sqrt{2t}$ and
$$g(u)=-e^u+u+t+1.$$
Since $t>0$, we always have $u<0$. To prove $g>0=g(0)$ (we now regard $g$ as a function of $u$ rather than $t$), we just need $\frac{dg}{du}<0$ and then we have $g(u)>g(0)=0$ when $u<0$.
$$\frac{dg}{du}=-e^u+1+\frac{dt}{du}$$
$$\frac{dt}{du}=\frac{1}{-\frac{1}{3}-\frac{1}{\sqrt{2t}}}=-\frac{3\sqrt{2t}}{3+\sqrt{2t}}$$
Thus, $$\frac{dg}{du}=-e^u+1-\frac{3\sqrt{2t}}{3+\sqrt{2t}}=-e^{-\frac{t}{3}-\sqrt{2t}}+1-\frac{3\sqrt{2t}}{3+\sqrt{2t}}.$$
Let $s=\sqrt{2t}>0$, we want to prove $$\frac{dg}{du}=-e^{-\frac{s^2}{6}-s}+1-\frac{3s}{3+s}<0,$$
which is equal to
$$f(s)=(s+3)e^{-s-\frac{s^2}{6}}+2s-3>0.$$
We notivce $f(0)=0$ and guess
$$f'(s)=e^{-s-\frac{s^2}{6}}+(s+3)e^{-s-\frac{s^2}{6}} (-1-\frac{s}{3})+2>0$$
$$6>(s^2+6s+6)e^{-s-\frac{s^2}{6}}$$
$$ln6>ln(s^2+6s+6)-s-\frac{s^2}{6}$$
$$s+\frac{s^2}{6}>ln(s+\frac{s^2}{6}+1)$$
For any $x\in \mathbb{R}$, we have $ln(x+1)<x$. And we have proved the inequality right.
