I was hoping someone could take the time to correct my proposed solution. Thanks.
Putnam 1976 A6:
Let $f : R \mapsto [-1,1]$ be twice differentiable and satisfy $(f(0))^2 + (f'(0))^2 = 4$. Show that $f(x_0) + f''(x_0) = 0$ for some $x_0 \in R$.
I have a (possibly overly complicated) solution different from the (much simpler) one found here https://prase.cz/kalva/, though it does use some of the same principles.
Solution:
Let $g(x) = (f(x))^2 + (f'(x))^2$. We have $g'(x) = 2f'(x)(f(x) + f''(x))$.
So, we are done if g attains a maximum on an open interval containing $0$. This is because if this maximum occurs at $x_0 \in (a,b)$ where $0\in(a,b)$, then $4 = g(0) \leq g(x_0) = (f(x_0))^2 + (f'(x_0))^2 \leq 1 + (f'(x_0))^2$ hence $\sqrt{3} \leq |f'(x_0)|$. Also, $g(x) \leq g(x_0)$ for all $x\in(a,b)$, so $\lim_{x\uparrow{x_0}}{\frac{g(x)-g(x_0)}{x-x_0}} \leq 0 \leq \lim_{x\downarrow{x_0}}{\frac{g(x)-g(x_0)}{x-x_0}}$ But $\lim_{x\uparrow{x_0}}{\frac{g(x)-g(x_0)}{x-x_0}} = \lim_{x\downarrow{x_0}}{\frac{g(x)-g(x_0)}{x-x_0}}$ by differentiability of $g$. So $g'(x_0) = 0$ and $f'(x_0) \neq 0$.
With this in mind, assume $g$ does not attain a maximum on any open interval containing $0$.
Then, intervals of the form $[a,b]$, where $a < 0 < b$ are real numbers, are such that $g$ attains a maximum at either $a$ or $b$ by the Extreme Value Theorem and our assumption. Call this maximizing input, either $a$ or $b$, $m(a,b)$. In the case where $g(a) = g(b)$ we pick arbitrarily.
Now fix $a < 0$ and suppose that for some $b^*>0$, $m(a,b^*)=a$. Now consider arbitrary $0<b<b^*$. We have $[a,b]\subseteq[a,b^*]$ so $g(a) \geq g(x)$ for all $x\in[a,b]$ so $m(a,b) = a$.
Now let $b^*_a=\sup\{b : a = m(a,b)\}$ for arbitrary $a < 0$.
Suppose that for all $a < 0$, $b^*_a = \infty$.Then for all $a < 0$ and arbitrary $b > 0$, $g(a) = g(m(a,b)) \geq 4 = g(0)$. Hence, $|f'(a)| \geq \sqrt{3}$ for all $a < 0$. By continuity of $f'$ we must have either $f'(a) \geq \sqrt{3}$ for all $a < 0$ or $f'(a) \leq -\sqrt{3}$ for all $a < 0$, in either case $f$ is unbounded, contradiction.
Now suppose there is some $a < 0$ such that $b^*_a$ is finite. Then for arbitrary $b^*_a < b$ we have $g(b) = g(m(a,b)) \geq 4 = g(0)$. Similar to the above case, we must have $|f'(b)| \geq \sqrt{3}$ for all $b^*_a < b$. By continuity of $f'$ we must have either $f'(b) \geq \sqrt{3}$ for all $b^*_a < b$ or $f'(b) \leq -\sqrt{3}$ for all $b^*_a < b$, in either case $f$ is unbounded, contradiction.
Thus $g$ must attain a maximum on an open interval containing $0$ and we're done.