2887
2021-05-24 09:27:42
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Observe that P(x) = x^4 P(1/x), one can conclude that if -p is a root of P, then so is -1/p.
Hence let -p, -q, -1/p, -1/q be the roots of P. Then P(x) = (x+p)(x+q)(x+1/p)(x+1/q).
Expanding the expression yields a = p + q + 1/p + 1/q and b = pq + p/q + q/p+1/pq+ 2.
Since -1/p < -1/2, we also have 1/2 < p < 2. Similarly, 1/2 < q < 2.
Let F(p,q) := 5a - 2b = -2pq + 5p + 5q -2p/q-2q/p + 5/p+5/q -2/pq -4.
It remains to show that F >= 8 on the rectangle 1/2 < p, q < 2.
Since F(p,q) = -(2q-1)(q-2)/q (p+1/p) + (5q^2-4q+5)/q,
for any given 1/2 < q < 2, F is minimized at p = 1 (plot p + 1/p). Similarly, for any given 1/2 < p < 2, F is minimized at q = 1. Therefore, F(p,q) >= F(1,q) >= F(1,1) = 8, for any 1/2 < p, q < 2.