Reverse QM-AM

We know that $x\ge 0, y\ge 0, c>0$ . Since the inequality is always stands when $x=y$ independently from $c$ , let's assume that $x\neq y$ .

Let $u=x+y, v=\left| x-y \right|$ . It's apparent that $u\ge v>0$ .

Then, ${ x }^{ 2 }+{ y }^{ 2 }=\frac { 1 }{ 2 } \left( 2{ x }^{ 2 }+2{ y }^{ 2 } \right) =\frac { 1 }{ 2 } \left( { x }^{ 2 }+2xy+{ y }^{ 2 }+{ x }^{ 2 }-2xy+{ y }^{ 2 } \right) =\frac { 1 }{ 2 } \left( { \left( x+y \right) }^{ 2 }+{ \left( x-y \right) }^{ 2 } \right) =\frac { 1 }{ 2 } ({ u }^{ 2 }+{ v }^{ 2 })$ .

Now, let's rewrite the inequality:

$\frac { u }{ 2 } +c\cdot v\ge \sqrt { \frac { 1 }{ 4 } \left( { u }^{ 2 }+{ v }^{ 2 } \right) }$

$\frac { 1 }{ 4 } { u }^{ 2 }+c\cdot uv+{ c }^{ 2 }\cdot { v }^{ 2 }\ge \frac { 1 }{ 4 } { u }^{ 2 }+\frac { 1 }{ 4 } { v }^{ 2 }$

$c\cdot uv+{ c }^{ 2 }\cdot { v }^{ 2 }\ge \frac { 1 }{ 4 } { v }^{ 2 }$

$c\cdot \frac { u }{ v } +{ c }^{ 2 }\ge \frac { 1 }{ 4 }$

Let $t=\frac { u }{ v } ,\quad t\ge 1$ :

${ c }^{ 2 }+t\cdot c-\frac { 1 }{ 4 } \ge 0$

The roots are ${ c }_{ 1,2 }=\frac { -t\pm \sqrt { { t }^{ 2 }+1 } }{ 2 }$ . Since $c>0$ , then $c\ge { c }_{ 2 }=\frac { -t+\sqrt { { t }^{ 2 }+1 } }{ 2 }$ .

$f\left( t \right) =\frac { -1+\sqrt { { t }^{ 2 }+1 } }{ 2 }$ is a descending function, meaning the maximum is reached at $t=1$ . That means $c\ge f\left( 1 \right) =\boxed { \frac { \sqrt { 2 } -1 }{ 2 } }$

Note: $t=1$ when $u=v$ meaning either $x$ or $y$ equals zero.