Is this true $(x-z)^2\leq e^{(x-y)^2}+e^{(y-z)^2}$ for all real numbers $x,y,z$? Is this true $(x-z)^2\leq e^{(x-y)^2}+e^{(y-z)^2}$ for all real numbers $x,y,z$?
We note that $(x-z)^2\leq (x-y)^2+(y-z)^2$ is not true for all real numbers $x,y,z$?
 A: Yes, the inequality holds true.
Indeed, using substitution $u=x-y$ and $v=y-z$ gives new statement
$$(u+v)^2\leq e^{u^2}+e^{v^2}\Rightarrow \underbrace{(u+v)^2-(e^{u^2}+e^{v^2})}_{(*)}\leq 0$$
To prove $(*)\leq 0$, using $(u+v)^2\leq 2(u^2+v^2)$ gives new bound
$$(*)\leq 2u^2-e^{u^2} + 2v^2-e^{v^2}$$
Now, knowing that the one variable function $f(w)=2w^2-e^{w^2}$ attains its absolute maximum at $w=\pm \sqrt{ln(2)}$ with value $f(\pm \sqrt{ln(2)})=2(ln(2)-1)<0$ gives us the desired inequality
A: Hint :
using Jensen's inequality we need to show :
$$2e^{\frac{\left(x-y\right)^{2}}{2}+\frac{\left(y-z\right)^{2}}{2}}>\left(x-z\right)^{2}$$
Taking the log we have :
$$\ln(2)+\frac{\left(x-y\right)^{2}}{2}+\frac{\left(y-z\right)^{2}}{2}\geq \ln((x-z)^2)$$
Or Setting $u=x-y,v=y-z$ as @Kafka we need to show :
$$F(u)=\ln(2)+\frac{\left(u\right)^{2}}{2}+\frac{\left(v\right)^{2}}{2}-\ln((u-v)^{2})>0$$
$$F'(u)=2/(u-v)+u$$
So we need to show :
$$G(u)=\ln(2)+\frac{\left(u\right)^{2}}{2}+\frac{\left(\frac{1}{2}\left(u+\sqrt{u^{2}+8}\right)\right)^{2}}{2}-\ln((\frac{1}{2}\left(u+\sqrt{u^{2}+8}\right)-u)^{2})>0$$
The derivative is :
$$G'(x)=\frac{1}{4}\cdot\frac{\left(2x^{2}+6x\sqrt{x^{2}+8}+16\right)}{\sqrt{x^{2}+8}}$$
I think now you can finish and as last remark we have $x=-1$ as roots for the derivative
