jmc

Since $x$ and $y$ are nonnegative, $x=a^2$ and $y=b^2$ for some nonnegative real numbers $a$ and $b.$ Then $$ab+c|a^2-b^2|\ge \frac{a^2+b^2}{2}.$$If $a=b,$ then both sides reduce to $a^2,$ and so the inequality holds. Otherwise, without loss of generality, we can assume that $a<b.$ Then the inequality above becomes $$ab+c(b^2-a^2)\ge \frac{a^2+b^2}{2}.$$Then $$c(b^2-a^2)\ge \frac{a^2+b^2}{2}-ab=\frac{a^2-2ab+b^2}{2}=\frac{(b-a{)}^2}{2},$$so $$c\ge \frac{(b-a{)}^2}{2(b^2-a^2)}=\frac{b-a}{2(b+a)}.$$We want this inequality to hold for all nonnegative real numbers $a$ and $b$ where $a<b.$

Note that $$\frac{b-a}{2(b+a)}<\frac{b+a}{2(b+a)}=\frac{1}{2}.$$Furthermore, by letting $a$ approach 0, we can make $\frac{b+a}{2(b-a)}$ arbitrarily close to $\frac{1}{2}.$ Hence, the smallest such real number $c$ is $\boxed{\frac{1}{2}}.$