How to show $13\left(x^{2}+1\right)+\sqrt{2\left(x^{4}+1\right)}+62 x-\frac{45\left(x^{2}-1\right)}{\ln x}>0$ for $x\in(0,1)$ I am doing a problem that can be reduced to proving the following inequality:

$$13\left(x^{2}+1\right)+\sqrt{2\left(x^{4}+1\right)}+62 x-\frac{45\left(x^{2}-1\right)}{\ln x}>0, x \in(0,1)
$$

But I don't know how to handle it. 
I tried to separate $\ln x$ from the function，because it is easier to differentiate. But I cannot handle the radical expression.
Any ideas?
 A: Here's my sketch proof:
For the following assume $x\in(0,1)$. First use the following lemma: 

$$\left(x^2+1\right)+\sqrt{2}\sqrt{x^4+1}>2x+2\left(\frac{x^4+1}{x^2+1}\right)$$

Proof:
$${\frac{d}{dx}\left(2(x^4+1)-\left\{2x+2\left(\frac{x^4+1}{x^2+1}\right)-(x^2+1)\right\}^2\right)\\=\frac{4(x-1)^5(x^4+2x^3+4x^2+4x+1)}{(x^2+1)^3}<0}$$
So plugging in $x=1$ to the original gives $4=4$ and we're done. 
Then it remains to show that

$$12\left(x^2+1\right)+2\left(\frac{x^4+1}{x^2+1}\right)+64x>45\left(\frac{x^2-1}{\ln
x}\right)$$

Proof: By algebraic manipulation, this is equivalent to 
$$\frac{2}{45}\ln x<\frac{x^4-1}{7x^4+32x^3+12x^2+32x+7}$$
So 
$$\frac{d}{dx}\left(\frac{2}{45}\ln x-\frac{x^4-1}{7x^4+32x^3+12x^2+32x+7}\right)\\[5px]=\frac{2(x-1)^6(49x^2+22x+49)}{45x(7x^4+32x^3+12x^2+32x+7)^2}>0$$
and plugging in $x=1$ to the equivalent form gives $0=0$ and we're done. 
A: $x \in(0,1)$ so
\begin{align}
13(x^{2}+1)&>0\\
\sqrt{2(x^{4}+1)}&>0\\
62 x&>0\\
-\frac{45(x^{2}-1)}{\ln x}&>0
\end{align}
A: It is obvious that 
$$\lim_{x\to  0}\frac{x^2-1}{\ln x}=0$$
now you should prove:
$$\lim_{x\to  1^-}\frac{x^2-1}{\ln x}=2^-$$
(use Hopital rule)
So:
$$13\Big((1)^{2}+1\Big)+\sqrt{2\Big((1)^{4}+1\Big)}+62(1)-{45}(2^-)>0$$
