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 $[1,e-1)$.

[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 $f'(x)>0$ ($x\geq 1$). 
