Find the maximum value of $a$ under the condition $2e\ln x\leq ax+b\leq \frac{1}{2}x^2+e$,where $a,b\in\mathbb{R}$,$x>0$

Question

Assume there exists $a,b\in \mathbb{R}$,such that $$2e\ln x\leq ax+b\leq \frac{1}{2}x^2+e$$ hold for $\forall x>0$.find the maximum value of $a$.
I guess the critical situation is $y=ax+b$ it the common tangent line of $f(x)=2e\ln x$ and $g(x)=\frac{1}{2}x^2+e$ where $e$ is the Euler constant.and assume the tangency point of $y$ and $f(x)$ is $(x_1,2e\ln x_1)$, the tangency point of $y$ and $g(x)$ is $(x_2,\frac{1}{2}x_2^2+e)$ then $$\begin{cases}x_2=a \\ \frac{2e}{x_1}=a \\ \frac{\frac{1}{2}x_2^2+e-2e\ln x_1}{x_2-x_1}=a\end{cases}$$ in order to find the maximum value of $a$,i want to get a function or equation of $a$, so I eliminate $x_1,x_2$ in above equations, then I get $$\ln\frac{2e}{a}=\frac{3}{2}-\frac{a^2}{4e}>\frac{3}{2}$$ then $$\frac{2e}{a}>e^{\frac{3}{2}}\Rightarrow a<\frac{2}{\sqrt{e}}$$ I can't go further more. I think the reason is i can't separate $a$ in $\ln\frac{2e}{a}=\frac{3}{2}-\frac{a^2}{4e}$.I only get a upper bound of $a$, not the maximum of $a$. and the critical situation is get by graph calculate. I can't find a strict proof.

Answer 1

When ax+b is tangent to the other two functions and b is the smallest, the value of a is the largest Let the tangent point be（x1，y1）（x2，y2）So the derivatives of the other functions is equal and equals to a Now we have x1=2e·1/x2 which means x₁x₂=2e So a=(x₁²/2+e)'=(x₁²/2+e-2elnx₂)/x₁-x₂ you can solve this equation and get X₂=e³/²-x₁²/4e（it is the power of e i dont know how to use the latex sorry…） recall that x₁x₂=2e now we have:x₁·e³/²-x₁²/4e=2e this is a transcendental function and we can guess by making x₁²/4e=1 which gives x₁=2√e so the maximum value of a is a=2√e