Proposed solution of problem OC179

Dear Editor of Crux Mathematicorum, I would like to submit the following solution of problem OC179

Find the maximum value of

|a^{2} − bc + 1| + |b^{2} − ac + 1| + |c^{2} − ab + 1|

where a, b, c are real numbers in [−2, 2].

19.

Proof Let’s suppose that each modulus is different from zero (so that |x| is twice differentiable) and consider f(a, b, c) = a^{2} + b^{2} + c^{2} − ab − bc − ca + 3. Since f_{a,a} = f_{b,b} = f_{c,c} = 2, it follows that f is convex and then the maximum is attained when each variable is +2 or −2. Since

f(2, 2, 2) = 3, f(2, −2, 2) = 19, f(2, −2, −2) = 19, f(−2, −2, −2) = 3

it follows that the maximum is 19.

Now let’s suppose that

a^{2} − bc + 1 = 0,  b^{2} − ac + 1 > 0,  c^{2} − ab + 1 > 0  (1)

We need to find the maximum of

b^{2} + c^{2} + 2 − a(b + c)

subject to (1). If b and c are free to span [−2, 2], we observe that the maximum is trivially 18; a value attained when b = c = 2, a = −2. Hence the maximum constrained by (1) is lower because −a = b = c = 2 doesn’t satisfy (1).

Now let’s assume

a^{2} − bc + 1 = 0,  b^{2} − ac + 1 > 0,  c^{2} − ab + 1 < 0  (2)

We need to find the maximum of

b^{2} − c^{2} + a(b − c) ≤ b^{2} + a(b − c)

subject to (2). If b and c are free to span [−2, 2], the maximum is 12; a value attained when b = c ± 2, a = ∓2. Hence the maximum constrained by (2) is lower because it doesn’t satisfy (2).

Now let’s assume

a^{2} − bc + 1 = 0,  b^{2} − ac + 1 < 0,  c^{2} − ab + 1 < 0  (3)

We need to find the maximum of

−b^{2} − c^{2} + a(b + c) − 2 ≤ a(b + c) ≤ 6

We don’t repeat here the above argument.

The other cases when b^{2} − ac + 1 = 0 and c^{2} − ab + 1 = 0 are analogous.

The conclusion is that 19 is the searched maximum.

Perfetti Paolo, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, 00133 Roma, Italy – email: perfetti@mat.uniroma2.it

Roma(Italy) 11/17/2015 Best regards Paolo Perfetti