Find the range of a variable with an inequality being true for any value of another variable.

Question

1) function $f(x)=\ln x-ax$ has two zeros $x_1, x_2$ and $x_10$,

$1+λ < \ln x_1+λ\ln x_2$ is true for any $x_1, x_2$, find the range of $λ$.

2) $\dfrac 2 {e^x+e^{-x}} \geq e^{ax^2}$ is true for any $x$, find the range of $a$.

Answer 1

Part (2)

$\dfrac{2}{e^{x}+e^{-x}}\geq e^{ax^2}$ gives $ax^2\leq\ln(\dfrac{1}{\cosh x})$ or $a\leq\dfrac{-1}{x^2}\ln(\cosh x)$.

$f(x)=\dfrac{-1}{x^2}\ln(\cosh x)$ is a real, continuous, even function, and increasing in $[0,\infty)$, because $\cosh x$ and $\ln x$ are increasing and $\dfrac{-1}{x^2}$ is increasing also. It's minimum occur in $x=0$. Since $$\lim_{x\to0}\dfrac{1}{x^2}\ln(\cosh x)=\dfrac{-1}{2}$$ Thus $\color{blue}{a\leqslant-\dfrac12}$.

Answer 2

I've come up with another solution for Part 2.

The inequality is the same as $ln2-ln(e^x+e^{-x})-ax^2\geq0$

Let $h(x)=ln2-ln(e^x+e^{-x})-ax^2$, $h(x)=0$, so

$h'(x)\leq0,x \to0^-;h'(x)\geq0,x\to0^+$

so $h''(0)\geq0$

$h''(x)=-\frac{4}{(e^x+e^{-x})^2}-2a$

$h''(x)_{min}=h''(0)=-1-2a\geq0$, $a\leq-\frac{1}{2}$

When $a\leq-\frac{1}{2}$, $h''(x)\geq0$

$h'(x)\leq0,x \leq0;h'(x)\geq0,x\geq0$

$h(x)_{min}=h(0)=0$

Thus $a\leq-\frac{1}{2}$ can be enough for the inequality.

For Part 1, I can so far tell $0\frac{1-lnx_1}{lnx_2-1}$.

But the most important is that I can't turn $x_1,x_2$ into one variable by some zooming so that I can get a function for maxmimum and minimum.