How to show $\sqrt{\log(1+x)}\leq 1+\log (x^2)$ for $x\geq 1$? Throughout the post, define $\log(x):=\log_e(x)$. Plotting the graph of the function
$$
f(x)=1+\log (x^2)-\sqrt{\log(1+x)},\quad x\geq 1,
$$
one can expect that $f(x)\geq 0$ for all $x\geq 1$.   



I'm looking for a proof of it. 

If one looks at its derivative:
$$
g(x)=f'(x)=\frac{-x + 4 \sqrt{\log(1 + x)} + 4 x \sqrt{\log(1 + x)}}{2 x (1 + x) \sqrt{\log(1 + x)}}
$$
then one ends up with showing
$$
4(x+1)\sqrt{\log(1+x)}\geq x,\quad x\geq 1
$$
which seems simpler than the original question. For $x\geq e-1$, this is immediate since $\log(1+x)\geq 1$. Still I would need to handle $[1,e-1)$.
 A: Suppose that $\sqrt{\ln(1+x)}> 1+\ln (x^2)$ for $x\geq 1$.
Square both sides to obtain $\ln(1+x)>1+4\ln x+4(\ln x)^2$.
Because you have $\ln (ex) > \ln(1+x)$ you also have $\ln(ex)=\ln e + \ln x=1+ \ln x > \ln (1+x) > 1+4\ln x+4(\ln x)^2$ for $ x \geq 1$ which becomes $4w^2+3w<0$ for $w \geq 0$ after substitution $w=\ln x$ and this is clearly false so your inequality is true.
So you do not even need derivatives.
A: [This is essentially Michael Rozenberg's answer. I would like to rephrase it a bit.]
It is mentioned in OP that it suffices to show that $f'(x)\geq 0$ for $x\geq 1$ since $f(1)>0$. On the other hand, when $x\geq 1$, 
$$
4(1+x)\sqrt{\ln(1+x)}-x\geq4(1+x)\sqrt{\ln2}-x>(4\sqrt{\ln 2}-1)x
>0,
$$
which implies that $f'(x)>0$ ($x\geq 1$). 
A: Just use $$4(1+x)\sqrt{\ln(1+x)}-x\geq4\sqrt{\ln2}(1+x)-x>0.$$
Hence, $f(x)\geq f(1)>0$ because $\ln2<1$ and we are done!
A: Proving your inequality is equivalent to proving that 
$$\log(1+x) \leq (1+\log(x^2))^2 = (1 + 2\log(x))^2 = 1 + 4\log(x) + 4(\log(x))^2.$$
Now define $f(x) = 1 + 4\log(x) + 4(\log(x))^2 - \log(1+x)$. Its derivative is 
$$f'(x) = \frac{3x+4}{x(x+1)} + \frac{8\log(x)}{x}.$$
Its clear that $f'(x) \geq 0$ for all $x \geq 1$, and since $f(1) = 1 - \log(2) \approx 1 - 0.69 >0$, we are done.
A: The quantity $\ln(1+x) - \ln(x)$ decreases in $(1, \infty)$, hence
$$\ln(1+x)\le \ln(2) + \ln(x)$$
The inequality
$$\ln(2) + \ln(x)\le (1 + 2\ln(x))^2$$
is easy to prove because the polynomial $p(u) = 4 u^2+3u+(1-\ln(2))$ is positive when $u\ge 0$ (because its minimum is at $u=-3/8<0$ and $p(0)>0$). Hence the result.
A: The square root functions is concave, therefore its graph is below
the tangent line at $x_0 = 1$:
$$
 \sqrt{x} \le \frac {1+x}{2} \quad \text{for } x \ge 0 \, .
$$
Using this inequality we get for $x \ge 1$:
$$
 \sqrt{\log(1+x)} \le \frac 12 \bigl(1 + \log(1+x) \bigr)
 = \frac 12 \bigl(1  + \log (1 + \frac 1x) + \log x \bigr) \\
 \le \frac 12 \bigl( 1 + \log 2 + \log x \bigr) 
 = \frac{1 + \log 2}{2} + \log (x^{1/2}) \\
  \approx 0.8466 +  \log (x^{1/2}) 
$$
which is a better estimate than the desired $1 + \log(x^2)$.
A: By setting $x=e^u$, it is enough to show that 
$$ \log(1+e^u) \leq (1+2u)^2 $$
holds for any $u\geq 0$. This is a pretty loose inequality, which can be proved by noticing first that $\log(1+e^u)\leq (1+2u)$, then noticing that $(1+2u)\leq (1+2u)^2$.
$\log(1+e^u)\leq 1+2u$ is equivalent to $1+e^u \leq e\cdot e^{2u}$ which clearly holds since the polynomial $p(x)=e x^2-x-1$ is increasing on $[1,+\infty)$ and $p(1)=e-2>0$.
