Functional inequality $\sqrt{f(x+y)}\leqslant\sqrt{f(x)}+\sqrt{f(y)}$ The following problem was posted on another forum but I wonder about the validity of its claim:
Let $f:\mathbb{R}\to [0, +\infty)$ satisfy the functional equation
$$f\left(\frac{x+y}{2}\right) +f\left(\frac{x-y}{2}\right) =\frac{f( x) +f( y)}{2}$$
Prove that
$$\sqrt{f( x+y)} \leqslant \sqrt{f( x)} +\sqrt{f( y)} ,\forall x,y\in [0, +\infty)$$
It seems the problem was motivated by the quadratic function $f(x)=x^2$. Indeed, it is easy to show that for any rational number $r$, we have $f(rx) = r^2f(x)$. So, if the function was continuous, we'd get the quadratic function and the inequality follows. However, no additional conditions are provided. My question is: is the statement of the problem correct and why, or if it is not correct, could you provide a counterexample?
 A: The claim is in fact true. Consider a function $ f : \mathbb R \to \mathbb R _ { 0 + } $ satisfying
$$ f \left ( \frac { x + y } 2 \right ) + f \left ( \frac { x - y } 2 \right ) = \frac { f ( x ) + f ( y ) } 2 \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $. Substitute $ x + y $ for $ x $ and $ x - y $ for $ y $ in \eqref{0} to get
$$ f ( x + y ) + f ( x - y ) = 2 f ( x ) + 2 f ( y ) \tag 1 \label 1 $$
for all $ x , y \in \mathbb R $. \eqref{1} is known as the quadratic functional equation (as it resembles the identity that holds for quadratic functions of the form $ f ( x ) = a x ^ 2 $ where $ a $ is a constant, with $ a \ge 0 $ here) or the parallelogram law functional equation (as it resembles the parallelogram law). You can take a look at the post "Parallelogram law functional equation: $ f ( x + y ) + f ( x - y ) = 2 \big( f ( x ) + f ( y ) \big) $" to check out some of the properties of the functions satisfying \eqref{1}. Although that post is about $ f : \mathbb R \to \mathbb R $, many of the things done there can be repeated here. The important property that is of use here is that there must exist a symmetric biadditive $ B : \mathbb R ^ 2 \to \mathbb R $ such that $ f ( x ) = B ( x , x ) $ for all $ x \in \mathbb R $. This lets us mimic the proof of the triangle inequality for inner product spaces by means of the Cauchy-Shwarz inequality. To do this, first note that since $ B $ is biadditive, it must be bilinear over $ \mathbb Q $. This is because fixing one of the arguments of $ B $ and varying the other, we get a function that satisfies the well-known Cauchy's functional equation, which you can find about by taking a look at "Overview of basic facts about Cauchy functional equation". Knowing this, one can define $ p : \mathbb Q \to \mathbb R $ with $ p ( t ) = f ( t x + y ) $ for all $ t \in \mathbb Q $, given any fixed $ x , y \in \mathbb R $, and observe that
\begin{align*}
p ( t ) & = f ( t x + y ) \\
& = B ( t x + y , t x + y ) \\
& = t ^ 2 B ( x , x ) + 2 t B ( x , y ) + B ( y , y ) \\
& = t ^ 2 f ( x ) + 2 t B ( x , y ) + f ( y )
\end{align*}
for all $ t \in \mathbb Q $. Therefore, $ p $ is in fact a quadratic polynomial on $ \mathbb Q $, and we can continuously extend it to $ \mathbb R $; i.e. defining $ q : \mathbb R \to \mathbb R $ with $ q ( s ) = s ^ 2 f ( x ) + 2 s B ( x , y ) + f ( y ) $ for all $ s \in \mathbb R $, $ q $ will be a continuous quadratic polynomial over $ \mathbb R $ with $ q | _ { \mathbb Q } = p $. Note that $ p $ only takes nonnegative values by definition, and since $ \mathbb Q $ is dense in $ \mathbb R $, $ q $ only takes nonnegative values, too. As this is only possible when the discriminant of $ q $ is nonpositive, we must have
$$ \bigl ( 2 B ( x , y ) \bigr ) ^ 2 - 4 f ( x ) f ( y ) \le 0 \text , $$
or equivalently
$$ | B ( x , y ) | \le \sqrt { f ( x ) f ( y ) } \text . \tag 2 \label 2 $$
Note that $ x $ and $ y $ were arbitrary, and therefore \eqref{2} holds for all $ x , y \in \mathbb R $. Finally, note that
\begin{align*}
f ( x + y ) & = B ( x + y , x + y ) \\
& = B ( x , x ) + 2 B ( x , y ) + B ( y , y ) \\
& = f ( x ) + 2 B ( x , y ) + f ( y ) \\
& \stackrel { \eqref{2} } \le \sqrt { f ( x ) } ^ 2 + 2 \sqrt { f ( x ) } \sqrt { f ( y ) } + \sqrt { f ( y ) } ^ 2 \\
& = \left ( \sqrt { f ( x ) } + \sqrt { f ( y ) } \right ) ^ 2
\end{align*}
for all $ x , y \in \mathbb R $, which by taking square roots of both sides proves the claim (even better, as it holds for all $ x , y \in \mathbb R $, rather than all $ x , y \in \mathbb R _ { 0 + } $).
