<s>[INST] 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 :)
 [/INST] Simple Verification
Once we know the constant is $\frac43$, the inequality is fairly simple.
$$
1+(x+y)^2\le\frac43\left(1+x^2\right)\left(1+y^2\right)\tag1
$$
is equivalent to
$$
\color{#C00}{1+4x^2y^2}+\color{#090}{x^2+y^2}\ge6xy\tag2
$$
which is true because the AM-GM says $\color{#C00}{1+4x^2y^2}\ge4xy$ and $\color{#090}{x^2+y^2}\ge2xy$. Equality is attained when $x=y=\frac1{\sqrt2}$.

Variational Argument To Get The Constant
Suppose we have have $\left(1+x^2\right)\left(1+y^2\right)$ fixed and we wish to maximize $1+(x+y)^2$. That is, for all $\delta x,\delta y$ so that
$$
2x\left(1+y^2\right)\delta x+2y\left(1+x^2\right)\delta y=0\tag3
$$
we also have
$$
2(x+y)(\delta x+\delta y)=0\tag4
$$
$(3)$, $(4)$, and orthogonality implies that there is a $\lambda$ so that
$$
2(x+y)=\lambda2x\left(1+y^2\right)\qquad\text{and}\qquad2(x+y)=\lambda2y\left(1+x^2\right)\tag5
$$
which implies
$$
x+\frac1x=y+\frac1y\tag6
$$
which means that
$$
y=x\qquad\text{or}\qquad y=\frac1x\tag7
$$
Let $\boldsymbol{y=\frac1x}$
$$
\begin{align}
\frac{1+(x+y)^2}{\left(1+x^2\right)\left(1+y^2\right)}
&=\frac{1+\left(x+\frac1x\right)^2}{\left(1+x^2\right)\left(1+\frac1{x^2}\right)}\\
&=\frac{1+\left(x+\frac1x\right)^2}{\left(x+\frac1x\right)^2}\\
&\le\frac54\tag8
\end{align}
$$
since $x+\frac1x\ge2$ (equality when $x=y=1$).
Let $\boldsymbol{x=y}$
$$
\begin{align}
\frac{1+(x+y)^2}{\left(1+x^2\right)\left(1+y^2\right)}
&=\frac{1+4x^2}{1+2x^2+x^4}\\
&=16\frac{1+4x^2}{16+32x^2+16x^4}\\[6pt]
&=\frac{16}3\frac1{\frac{1+4x^2}3+2+\frac3{1+4x^2}}\\
&\le\frac43\tag9
\end{align}
$$
since $\frac{1+4x^2}3+\frac3{1+4x^2}\ge2$ (equality when $x=y=\frac1{\sqrt2}$).
Therefore, we get that
$$
1+(x+y)^2\le\frac43\left(1+x^2\right)\left(1+y^2\right)\tag{10}
$$
</s>