2014/079) Suppose that a, b, c are sides of of a triangle. Prove that:

$a^2(b + c - a) + b^2(c + a - b) + c^2(a + b - c) \le 3abc$

proof:

For the angles of any triangle always
$\cos\ A + \cos\ B + \cos\ C ≤ 1.5$

further

$c² = a² + b² - 2ab\ cos\ C$
hence $c³ = a²c + b²c - 2abc\cos\ C$

$a²c + b²c - c³ = 2abc\cos\ C$
similarly
$ab² + ac² - a³ = 2abc\cos\ A$
$bc² + a²b - b³ = 2abc\cos\ B$

adding we get


$a²c + b²c - c³ + ab² + ac² - a³ + bc² + a²b - b³ = 2abc (\cos\ A + \cos\ B + \cos\ C)$

as

$\cos\ A + \cos\ B + \cos\ C ≤ 1.5$

$a²c + b²c - c³ + ab² + ac² - a³ + bc² + a²b - b³ ≤ 3abc$
or $(a²b + a²c - a³) + (b²c + ab² - b³) + (ac² + bc² - c³) ≤ 3abc$

or $a²(b + c - a) + b²(c + a - b) + c²(a + b - c) ≤ 3abc$
hence proved