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

Question

Find the inequality with the best possible $constant$

1. 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.
2. 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.

Answer 1

For example.

Let $x$ and $y$ be non-negative numbers such that $x^2+y^2\leq\frac{2}{15}.$ Prove that: $$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1.5}{1+ xy}\leq \frac{3.5}{1+ \left ( \frac{x+ y}{2} \right )^{2}},$$ where $\frac{2}{15}$ is a best such constant.

It's interesting that:

For any non-negatives $x$ and $y$ the following inequality is true. $$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{2}{1+ xy}\geq \frac{4}{1+ \left ( \frac{x+ y}{2} \right )^{2}}$$

By the way, the last inequality is true for any reals $x$ and $y$ such that $xy+1>0.$

Answer 2

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

$$ k_n = \frac{2n}{3n+4} $$