How prove $(2+5x)\ln{x}-6(x-1)>0.\forall x>1$ let $x>1$ show that:
$$(2+5x)\ln{x}-6(x-1)>0.\forall x>1$$
Let
$$f(x)=(2+5x)\ln{x}-6(x-1),~~~f'(x)=\dfrac{2}{x}+5\ln{x}-1$$
since
$f(1)=1$.so it must prove
$$f'(x)=\dfrac{2}{x}+5\ln{x}-1>0?$$
 A: Let $f(x)=\ln{x}-\frac{6(x-1)}{5x+2}$.
Hence,
$$f'(x)=\frac{1}{x}-6\cdot\frac{5x+2-5(x-1)}{(5x+2)^2}=\frac{1}{x}-\frac{42}{(5x+2)^2}=$$
$$=\frac{25x^2-22x+4}{x(5x+2)^2}=\frac{25x^2-25x+3x+4}{x(5x+2)^2}>0$$
for all $x>1$.
Thus, $f(x)>f(1)=0$ and we are done!
A: You are doing it right. First, $f$ is continuous and $f(1)=0$. Next, $f'(1)=1$ and $f'(x)>0$ for $x\ge 1$. Thus, $f(x)>0$ for $x>1$.
We have to show that $f'(x)>0$ for $x\ge 1$. At the interval $1<x<2$ we have $5\ln x>0$ and $\frac{2}{x}>1$, thus $f'(x)>0$. Next consider $x\ge 2$. Here we have $\frac{2}{x}>0$ and $5\ln x> 3$, thus $f'(x)>0$.
A: write your inequality for $$x>1$$ in the form
$$\ln(x)-\frac{6(x-1)}{2+5x}>0$$ and define $$f(x)=\ln(x)-\frac{6(x-1)}{2+5x}$$
and compute the derivative with respect to $x$
A: In what follows we assume $x>1$.

We are supposed to prove $$\log x>\frac{6(x-1)}{5x+2}\tag{1}$$ We already know the fundamental inequality $$\log x>\frac{x-1}{x}\tag{2}$$ Replacing $x$ by $\sqrt{x} $ we see that $$\log x>\frac{2(\sqrt{x}-1)}{\sqrt{x}}\tag{3}$$ and thus our job is done if we show that $$\frac{\sqrt{x}-1}{\sqrt{x}}>\frac{3(x-1) }{5x+2}$$ or $$\frac{1}{\sqrt{x}} >\frac{3(\sqrt{x}+1)}{5x+2}$$ Putting $x=t^2$ we see that this equivalent to $$2t^{2}-3t+2>0$$ which is true because the discriminant is negative. Note that the inequality $(2)$ is not sufficient to prove $(1)$, but can be strengthened into $(3)$ via a simple substitution and that suffices here. 
