Problem: Let $a,b,c$ be real numbers in the interval $[0,1]$
with $a+b,b+c,c+a \ge 1$.  Prove that
\[ 1 \le (1-a)^2 + (1-b)^2 + (1-c)^2
  + \frac{2\sqrt2 abc}{\sqrt{a^2+b^2+c^2}}. \]

Solution: The following approach is due to Ashwin Sah.

We will prove the inequality for any $a$, $b$, $c$
the sides of a possibly degenerate triangle
(which is implied by the condition),
ignoring the particular constant $1$.
Homogenizing, we instead prove the problem
in the following form:
\begin{claim*}
  We have
  \[ k^2 \le (k-a)^2 + (k-b)^2 + (k-c)^2
    + \frac{2\sqrt2 abc}{\sqrt{a^2+b^2+c^2}} \]
  for any $a$, $b$, $c$, $k$
  with $(a,b,c)$ the sides of a
  possibly degenerate triangle.
\end{claim*}
\begin{proof}
  For any particular $(a,b,c)$ this is a quadratic in $k$
  of the form $2k^2 - 2(a+b+c)k + C \ge 0$;
  thus we will verify it holds for $k = \half(a+b+c)$.

  Letting $x = \half(b+c-a)$ as is usual,
  the problem rearranges to
  In that case, the problem amounts to
  \[ (x+y+z)^2 \le x^2+y^2+z^2
    + \frac{2(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}} \]
  or equivalently
  \[ x^2+y^2+z^2+xy+yz+zx
    \le \left( \frac{(x+y)(y+z)(z+x)}{xy+yz+zx} \right)^2. \]
  To show this, one may let $t = xy+yz+zx$,
  then using $(x+y)(x+z) = x^2+B$ this becomes
  \[ t^2(x^2+y^2+z^2+t)
    \le (x^2+t)(y^2+t)(z^2+t) \]
  which is obvious upon expansion.
\end{proof}

\begin{remark*}
  The inequality holds actually for
  all real numbers $a$, $b$, $c$,
  with very disgusting proofs.
\end{remark*}