Find the inequality with the best possible $k= constant$ (with the condition $x^{2}+ y^{2}\leq k$). 
Find the inequality with the best possible $constant$


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*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{7}$. Prove that
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1}{1+ xy}\leq \frac{3}{1+ \left ( \frac{x+ y}{2} \right )^{2}}$$
where $constant= \frac{2}{7}$ is the best possible.

*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{5}$. Prove that
$$\frac{1}{\sqrt{1+ x^{2}}}+ \frac{1}{\sqrt{1+ y^{2}}}+ \frac{1}{\sqrt{1+ xy}}\leq \frac{3}{\sqrt{1+ \left ( \frac{x+ y}{2} \right )^{2}}}$$
where $constant= \frac{2}{5}$ is the best possible.

They are my two examples. I'm looking forward to seeing more many inequalities alike. Thanks for all your nice comments.

Hint.
After rotating CCW the inequality
$$
\frac{3}{\frac{1}{4} (a+b)^2+1}-\frac{1}{a^2+1}-\frac{1}{a b+1}-\frac{1}{b^2+1}\ge 0
$$
we have
$$
\frac{6}{y^2+2}-\frac{2}{-x^2+y^2+2}-\frac{2}{(x+y)^2+2}-\frac{2}{(x-y)^2+2}\ge 0\ \ \ \ \ \ (1)
$$
this has at the equality, the trace in blue shown in the figure below.
Thus to guarantee $x^2+y^2 = k$ with maximum $k$, the circle should be internally tangent to this curve. This could be easily calculated by doing in $(1)$, $x=0$ but this curve has immersed a double zero so we should proceed with
$$
\lim_{x\to 0}\frac{\frac{6}{y^2+2}-\frac{2}{-x^2+y^2+2}-\frac{2}{(x+y)^2+2}-\frac{2}{(x-y)^2+2}}{x^2} = \frac{4-14 y^2}{\left(y^2+2\right)^3}
$$
and thus solving $4-14 y^2=0$ we obtain the value for $k = \frac 27$

Applying the same procedure in the case of
$$
\frac{3}{\sqrt{\frac{1}{4} (a+b)^2+1}}-\frac{1}{\sqrt{a^2+1}}-\frac{1}{\sqrt{a b+1}}-\frac{1}{\sqrt{b^2+1}}\ge 0
$$
we obtain
$$
ineq=\frac{3}{\sqrt{y^2+2}}-\frac{1}{\sqrt{-x^2+y^2+2}}-\frac{1}{\sqrt{(x+y)^2+2}}-\frac{1}{\sqrt{(x-y)^2+2}}\ge 0
$$
and
$$
\lim_{x\to 0}\frac{ineq}{x^2} = \frac{2-5 y^2}{\sqrt{2} \left(y^2+2\right)^{5/2}}
$$
thus obtaining $k = \frac 25$ and also in the case of
$$
\frac{3}{\sqrt[3]{\frac{1}{4} (a+b)^2+1}}-\frac{1}{\sqrt[3]{a^2+1}}-\frac{1}{\sqrt[3]{a b+1}}-\frac{1}{\sqrt[3]{b^2+1}}\ge 0
$$
analogously we obtain
$$
k = \frac{6}{13}
$$
etc.
NOTE
for the $\sqrt[n]{\cdot}$ case we have
$$
k_n = \frac{2n}{3n+4}
$$
