Inequality, finding Constant Can anyone help me with this Math Olympiad Task from 2007 from Germany? I want to find the smallest $C$, such that for every $x,y \in \mathbb{R}$ the inequality:
$$
1+(x+y)^2 \leq C(1+x^2)(1+y^2)
$$
holds.
I know that I have to maximize the function
$$a(x, y) = \frac{1+(x+y)^2}{(1+x^2)(1+y^2)}$$
which gives me $\frac{4}{3}.$
P.S.: Thanks for the help y'all :)

Claim: We will show that $ 1  + (x+y)^2 \leq \frac{4}{3} ( 1+x^2)(1+y^2)$.
Proof: By expanding, WTS
$$ 4x^2y^2 + x^2 + y^2 + 1 \geq 6 xy. $$
This is true by applying AM-GM creatively:

 $ 4x^2 y^2 + 1 \geq  4 |xy | \geq 4 xy $.
$x^2 + y^2 \geq 2 | xy | \geq 2xy$.

Equality holds iff $ 2xy = 1$,  $ x = y$ and $ xy \geq 0$, which gives the solution set  $ x = y = \pm \frac{1}{ \sqrt{2} } $.
This solution sets also what that $\frac{4}{3}$ is the smallest possible value of $C$.

Note:

*

*As to how one can guess the value of $C$, we use the huge wishful thinking simplification that $x=y$, and have the quadratic equation in $t = x^2$ of
$$ C t ^2 + ( 2 C - 4 )t  + (  C - 1 ) \geq 0 \quad t \geq 0. $$
To satisfy this, we require
A)  If $ t = - \frac{ 2C-4}{C } \geq 0 $, then $f(t) \geq 0$ $\Rightarrow$ if $ 0 < C < 2 $, then $ C \leq \frac{4}{3}$, so the solution set is $ 2 > C \geq \frac{4}{3} $.
B)  Else if $ t = - \frac{ 2C-4}{C } \leq 0$, then $ f(0) \geq 0$ $\Rightarrow$ If $ t < 0$ or $ t > 2$, then $  C > 1 $. So the solution set is $ C \geq 2$.
Hence, $ C \geq \frac{4}{3}$, so the minimum value to try is $ C = \frac{4}{3}$.
Note that this might not work because the equality condition might not occur at $ x = y$.

