how to prove using elementary comparisons that $x \log x \leq \log(2x^2-3x+2)$?

Question

I am trying to show that $x \log x \leq \log(2x^2-3x+2)$ on $[1,2]$. The things is that these two functions are pretty close to each other:

so I have a hard time finding a simple function which trivially lies between the two.

Answer 1

Let $f(x)=\ln(2x^2-3x+2)-x\ln x$. We want to show that $f(x)\ge 0$ on $[1,2]$.

We have $$f'(x)=\frac{4x-3}{2x^2-3x+2}-(\ln x+1)$$ $$f''(x)=\frac{-4x^4+4x^3-5x^2+11x-4}{x (2x^2-3x+2)^2}$$ Here, let $g(x)=-4x^4+4x^3-5x^2+11x-4$. Then, $$g'(x)=-16x^3+12x^2-10x+11,\quad g''(x)=-48\left(x-\frac 14\right)^2-7\lt 0$$ So, $g'(x)$ is decreasing. Since $g'(1)=-3\lt 0$, we know that $g'(x)\lt 0$ on $[1,2]$.

Hence, $g(x)$ is decreasing on $[1,2]$. Since $g(1)=2,g(2)=-34$, there exists only one real number $\alpha$ such that $g(\alpha)=0$ and $1\lt \alpha\lt 2$.

So, we know that $f'(x)$ is increasing on $[1,\alpha)$ and is decreasing on $(\alpha,2]$. It follows from $f'(1)=0,f'(2)=\frac 14-\ln 2\lt 0$ that there exists only one real number $\beta$ such that $f'(\beta)=0$ and $\alpha\lt \beta\lt 2$.

Hence, it follows from this that $f(x)$ is increasing on $(1,\beta)$ and is decreasing on $(\beta,2]$.

Since $f(1)=f(2)=0$, we know that $f(x)\ge 0$ on $[1,2]$.

Answer 2

We have to compare $(1+x)\log(1+x)$ and $\log(1+x+2x^2)$ on $I=[0,1]$.

That is the same as comparing: $$ f(x)=(1+x)\log(1+x)-2x\log(2),\quad g(x)=\log(1+x+2x^2)-2x\log(2)\tag{1} $$ that are both vanishing functions on the endpoint of $I$. $f(x)$ is a convex function, hence $f(x)\leq 0$ on $I$. Since $g\left(\frac{1}{2}\right)=0$ but $g$ is positive on $\left(\frac{1}{2},0\right)$, it is enough to prove $f(x)\leq g(x)$ on $J=\left[0,\frac{1}{2}\right]$. We have $f(x)\leq (1-2\log 2)x+\frac{x^2}{2}$ over such interval, hence it is enough to prove: $$\forall x\in\left[0,\frac{1}{2}\right],\quad \log(1+x+2x^2)\geq x+\frac{x^2}{2}.\tag{2}$$ The last inequality is easy to prove if we notice that $\frac{\log(1+x+2x^2)}{x}$ is a concave function on $J$, or that $x+\frac{3x^2}{2}-\frac{5x^3}{3}$ is between the two terms given by $(2)$.