let $f(x)$ be two derivable on $R$,give the two postive numbers $A,B$ and such $$[f(x)]^2\le A$$ $$[f'(x)]^2+[f''(x)]^2\le B$$ show that $$[f(x)]^2+[f'(x)]^2\le \max{(A,B)},\forall x\in R$$
I think maybe well know inequality： $$|f'(x)|\le2\sqrt{|f''(x)||f(x)|}\le |f''(x)|+|f(x)|$$ $$\Longrightarrow [f'(x)]^2\le (|f''(x)|+|f(x)|)^2\le\dfrac{1}{2}([f''(x)]^2+[f(x)]^2)$$
if $A\ge B$ then I can't $$[f(x)]^2+[f'(x)]^2\le A$$
Maybe this idea can't usefulll?
Let $g(x)=f(x)^2+f'(x)^2$. Then $g'(x)=2f'(x)(f(x)+f''(x))$. It is sufficient to verify $g(x)\le\max\{A,B\}$ at critical points and as $x\to \pm\infty$.. At critical points of $g$ we have
As $x\to \infty$ beyond critical points, $g'(x)$ has constant sign, hence $f'(x)$ does not cross $0$ and so $f$ is monotonic. Since $f$ is bounded, we conclude that $a:=\lim_{x\to\infty}f(x)$ exists. Let $\epsilon>0$ be given, wlog. $\epsilon<\frac12$. Then there exists $x_0$ such that both $|f(x)-a|<\epsilon$ and $|f^2(x)-a^2|<\frac12\epsilon$ for all $x>x_0$. Then for any $x>x_0$ we have $f(x+2)-f(x)=2f'(\xi)$ for some $\xi\in(x,x+2)$. For this $\xi$ we conlcude $|f'(\xi)|<\epsilon$ and then $g(\xi)=f(\xi)^2+f'(\xi)^2$ differs from $a^2$ by less than $\epsilon$. As it lacks critical points, we must have $g(x)\to a^2\le A$. The same argument works as $x\to-\infty$.
