Show that $x^3 - 2 x^2 +\log(1+x)(x(3x+4) -2(1+x)^2 \log(1+x))$ is positive

Question

I want to show that (the following just gets rid off large brackets) $$x^3 - 2 x^2 +x(3x+4)\log(1+x)-2(1+x)^2 \log^2(1+x)>0, \ \ \mbox{for}\ \ x\in(0,\infty).$$

My attempt: Transfer all negative terms to the other side of the inequality. Evaluate the both sides at different limits of $x$, and see that both sides converge to $0$ for $x \to 0$ and increase infinitely for $x \to \infty$. Take the derivative of both sides with respect to $x$ to demonstrate that the left side increases faster than the right side. Burn out at that point because showing that the last point holds I need to engage into a similar task as proving the original inequality.

Answer 1

Solve with respect to $\log(1+x)$ to obtain \begin{equation} \frac{3x^2 + 4x - x^2\sqrt{8x+9}}{4(1+2x+x^2)}<\log(1+x)<\frac{3x^2 + 4x + x^2\sqrt{8x+9}}{4(1+2x+x^2)}, \ \ \mbox{for}\ \ x>0. \tag{*} \end{equation} Start with the first inequality. Observe that both sides of the inequality go to zero when $x\to 0$. Also it is easy to check the LHS (containing the fraction term) is increasing starting from zero and then is decreasing in $x$, while the RHS (containing $\log$) is increasing in $x$ and goes to infinity when $x\to \infty$. Therefore, the inequality holds if the RHS is increasing faster than the LHS, or formally (taking derivatives, respectively): $$\frac{-10 x^2 - 2 x^3 + 2\sqrt{9 + 8 x} + x (-9 + \sqrt{9 + 8 x})}{2(1+x)^3\sqrt{9+8x}} < \frac{1}{1+x}.$$ Proving the last inequality holds should not be a problem. Hence, the first inequality of $(*)$ is true.

Do similar steps to show that the second inequality in $(*)$ holds.

Answer 2

By solving with respect to $\log(1+x)$ the given inequality is equivalent to $$\forall x>0,\qquad \log(1+x)\leq \frac{4x+3x^2+x^2\sqrt{9+8x}}{4(x+1)^2} \tag{1}$$ which is pretty accurate, more accurate than $\log(1+x)\leq \sqrt{x+1}-\frac{1}{\sqrt{x+1}}$ which can be derived by applying the Cauchy-Schwarz inequality to $\log(1+x)=\int_{0}^{x}\frac{dt}{1+t}$. On the other hand, if we apply $\int_{0}^{x}(\ldots)\,dt$ to both sides of $$ \log(1+t) \leq \sqrt{t+1}-\frac{1}{\sqrt{t+1}} $$ and rearrange we get $$ \log(1+x) \leq \frac{\frac{4}{3}+x-\frac{2}{3}(2-x)\sqrt{x+1}}{x+1}\tag{2} $$ which is sharper than $(1)$ for any $x\geq 2$. It remains to prove $(1)$ over $(0,2)$, and that can be done through Taylor series, for instance. I hope you appreciate the trick for producing sharp inequalities: start with CS, then integrate and rearrange multiple times.

Answer 3

COMMENT.-I give here a proof with graph of functions. Unlike, for example, the inequality $x ^ 2 + 5 \gt x$ in which the graphs show with quite clarity the inequality, for the proposed one the difference is very small in the neighborhood of $0$, to such an extent that the graphs are confused in a single portion of the graphs of $f(x)=x^3 - 2 x^2 +x(3x+4)\log(1+x)$ and $g(x)=2(1+x)^2 \log^2(1+x)$.

Let $F(x)$ be defined by $F(x)=f(x)-g(x)$ so

$$F(x)=x^3 - 2 x^2 +x(3x+4)\log(1+x)-2(1+x)^2 \log^2(1+x)>0, \ \ \mbox{for}\ \ x\in(0,\infty).$$

We have $F(0)=0,\space F(0.1)\approx 0.0000000238099,\space F(0.2)\approx 0.0000013200437$ which suggests that the function $F$ is increasing and because $F(0)=0$ it follows that if the derivative $F'(x)$ is positive then it is proved that $F(x)\gt 0$. I don't know if solving the inequality $$F'(x)=\frac{3x^3+2x^2}{1+x}-4(1+x)\log^2(1+x)+2x\log(1+x)\gt 0$$ it is of less or equal difficulty than $F (x)\gt 0$ but here it is about seeing the evidentiary graphs. In fact, contrary to the impossibility of seeing graphically that $f (x)\gt g(x)$ (i.e. $F(x)\gt 0$) for the derivative $F '(x)$ it is clearly seen that it is positive therefore $F(x)\gt 0$.