Prove that if $f^2(x+y)+f^2(x-y)=2f^2(x)+2f^2(y)$ then $f(x+y) \leq f(x)+f(y)$ for all $x,y \in \mathbb R$ Let $f: \mathbb{R} \rightarrow [0, \infty)$ be a function that:
$$f^2(x+y)+f^2(x-y)=2f^2(x)+2f^2(y) \forall x,y \in \mathbb{R}$$
Prove that $f(x+y) \leq f(x)+f(y) \forall x,y \in \mathbb{R}$
This problem is from this AOPS link.
Let $x=y=0$,then $f(0)=0$. then let $x=y$,we have
$$f^2(2x)=4f^2(x)\Longrightarrow f(2x)=2f(x)$$
let $x=2y$,then we have
$$f^2(3y)+f^2(y)=2f^2(2y)+2f^2(y)=10f^2(y)\Longrightarrow f(3x)=3f(x)$$
let $x=3y$ then we have $f(4x)=4f(x)$,and use induction we have
$$f(kx)=kf(x),\forall k\in N^{+}$$ then I can't it
 A: This is a standard result in linear algebra. Here I write a solution accessible to high school students, as the question is originally posted as a high school contest question.
You have already shown that $f(kz) = kf(z)$ for all $k \in \Bbb Z_{\geq 0}$ and all $z \in \Bbb R$.
Let $x, y$ be any real number.
If $f(x) = f(y) = 0$, then we have $f(x + y)^2 + f(x - y)^2 = 0$ and hence $f(x + y) = 0$.
Thus in the following we assume without loss of generality that $f(y) \neq 0$.
We write $p = \frac 1 2(f(x + y)^2 - f(x)^2 - f(y)^2)$ and prove by induction on $m + n$ that $$f(mx + ny)^2 = m^2f(x)^2 + 2mnp + n^2f(y)^2\tag{*}$$ for all nonnegative integers $m, n$.
The claim is clearly true for $m + n \leq 2$. When $m + n \geq 3$, we assume without loss of generality that $m \geq 2$ and use the original functional equation with $x, y$ replaced with $(m - 1)x + ny$ and $x$ to finish the inductive step.
We have thus proved $(*)$ for all nonnegative integers $m, n$. Using the original functional equation with $x, y$ replaced with $mx, ny$, we see that $(*)$ is in fact true for all $m \in \Bbb Z_{\geq 0}$ and all $n \in \Bbb Z$.
Applying the identity $f(kz) = kf(z)$, we have $f(x + \lambda y)^2 = f(x)^2 + 2\lambda p + \lambda^2f(y)^2$ for all rational number $\lambda$.
Thus we know that the quadratic polynomial $Q(\lambda) = f(y)^2 \lambda^2 + 2p\lambda + f(x)^2$ takes nonnegative values for all rational number $\lambda$.
This is only possible when its discriminant is not positive. Therefore we have $p^2 \leq f(x)^2f(y)^2$, or equivalently, $|p| \leq f(x)f(y)$.
Consequently, we have $f(x + y)^2 = f(x)^2 + 2p + f(y)^2 \leq (f(x) + f(y))^2$ and hence $f(x + y) \leq f(x) + f(y)$.
A: For an inner product space, we can define $\|x\|=\sqrt{(x,x)}$ to be a norm, which fulfill triangular inequallity: $\|x+y\|\le \|x\|+\|y\|$ by Cauchy-Schwartz inequallity. We have parallelogram identity:
$$
\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2
$$
And polarization identity (in real space):
$$
(x,y)=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2)
$$
A proposition in functional analysis is that, norm may not be induced by an inner product, but once your norm satisfies parallelogram identity, you can define an inner product by polarization identity. And this inner product will exactly induce your original norm.
So try to define an inner product by $f$, verify it is an inner product, and then use property of inner product to prove the triangular inequallity.

It takes me a lot of time to consider the following question: If the space we're working is $\mathbb{R},$ then if $f$ is a (semi-)norm, won't we directly have $$f(x+y)=f(x)+f(y)?$$
So we shall regard $\mathbb{R}$ as a $\mathbb{Q}$-vector space, and $f$ will be a semi-norm following $\mathbb{Q}$-linearity. You will have the $\mathbb{Q}$-linearity of your $(-,-)$ by parallelogram identity. Prove the (semi) inner product satisfies Cauchy-Schwartz inequality, using $(x+\lambda y,x+\lambda y)\ge 0,$ $\lambda\in\mathbb{Q}.$ You will need to take limit of $\lambda\rightarrow \eta\in \mathbb{R}.$ However it doesn't need the continuity of $f.$ It only uses the topology in $\mathbb{R}$.

Above are all my thinking and backgrounds. The following is the answer to this question, which doesn't need many techniques.
You already have: $$f(0)=0, \qquad f(kx)=kf(x),\quad\forall k\in \mathbb{Q}$$
Let $(x,y)=\frac{1}{4}(f(x+y)^2-f(x-y)^2).$ Then it has the following properties:

*

*$(x,x)=f(x)^2\ge 0,$ and equal to zero when $x=0$;


*$(x,y)=(y,x),$ as $f(x-y)^2=f(y-x)^2$;


*$(x+y,z)=(x,z)+(y,z).$ This is because:
$$
\begin{aligned}
(x,z)+(y,z)&=\frac{1}{4}(f(x+z)^2-f(x-z)^2+f(y+z)^2-f(y-z)^2)\\
&=\frac{1}{2}(f(\frac{x+y}{2}+z)^2+f(\frac{x-y}{2})^2-f(\frac{x+y}{2}-z)^2-f(\frac{x-y}{2})^2)\\
&=2(\frac{x+y}{2},z)\\
(0,z)&=\frac{1}{4}(f(z)^2-f(-z)^2)=0\\
(x+y,z)+(0,z)&=2(\frac{x+y}{2},z)
\end{aligned}
$$


*$\forall k\in \mathbb{Q},$ $(kx,y)=k(x,y).$ Denote $F(t)=(tx,y),$ then by property 3, $F(t_1+t_2)=F(t_1)+F(t_2).$ This shows that $F(t)=tF(1),$ $\forall t\in \mathbb{Q}.$
Then we verify that $(-,-)$ satisfies Cauchy-Schwartz inequality:
$$(x,y)^2\le (x,x)(y,y).$$
Use $(x+\lambda y,x+\lambda y)\ge 0,$ $\lambda \in \mathbb{Q},$
$$
(x,x)+2\lambda(x,y)+\lambda^2(y,y)\ge 0.
$$
Let $\lambda\rightarrow -\frac{(x,y)}{(y,y)},$ the inequality keeps, so
$$
(x,x)-\frac{(x,y)^2}{(y,y)}\ge 0\Rightarrow (x,y)^2\le (x,x)(y,y).
$$
Finally, we have
$$
\begin{aligned}
f(x+y)^2&=(x+y,x+y)\\
&=(x,x)+2(x,y)+(y,y)\\
&\le(x,x)+2\sqrt{(x,x)(y,y)}+(y,y)\\
&=(\sqrt{(x,x)}+\sqrt{(y,y)})^2=(f(x)+f(y))^2
\end{aligned}
$$
Triangular inequality is proved.
So the main step is to prove $\mathbb{Q}$-linearity of the inner product and Cauchy-Schwartz inequality. Retain this two step, you can find equivalent proof by only using $f,$ without defining $(-,-).$ However, I think it's pretty  concise to prove it with the help of $(-,-).$
