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 :)
 A: Since it must be true for all $x,y$ then it must be true also for $x=y$ and we get $$Ct^2-4t+3\geq 0$$ where $t=x^2+1\geq 1$.
Clearly $C>0$ so we can rewrite it like this $$(Ct-2)^2 -4+3C\geq 0$$
If $C<4/3$ then for $t={2\over c} >1$
we get $$-4+3C\geq 0 \implies C\geq {4\over 3}$$

Now, the game is not over yet. You must prove that this value actually works for all $x,y$:
$$3+3x^2+6xy+3y^2\leq 4+4x^2+4y^2+4x^2y^2$$  which is equivalent to $$0\leq (x-y)^2+(2xy-1)^2$$ which is obviously true.
A: 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
$$
1+(x+y)^2\le\frac43\left(1+x^2\right)\left(1+y^2\right)\tag{10}
$$
A: The command of Mathematica
Resolve[ForAll[{x, y}, 1 + (x + y)^2 <= c*(1 + x^2)*(1 + y^2)], Reals]

answers $c\geq \frac{4}{3}$
Addition. and the command of Mathematica
Maximize[(1 + (x + y)^2)/(1 + x^2)/(1 + y^2), {x, y}]


$\left\{\frac{4}{3},\left\{x\to -\frac{1}{\sqrt{2}},y\to -\frac{1}{\sqrt{2}}\right\}\right\}$

shows how to derive it by hand.
A: 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$.

