Page 21 - MATINF Nr. 8

3.3 Lower bounds 21

3.3 Lower bounds

√
Now, let us find the best numerical lower bound of the product under consideration for k ≥ 3 ,
3
and over all non-obtuse-angled triangles ABC. We will also extend the result to nonnegative
real numbers x, y, z.
√
2
Just like before, denote p = 3t +1 , where t is fixed, t ≥ 3 . According to Theorem 3 from [1],
2t 3
the minimum value of r is attained when (x, y, z) is of the form
1 − t
Å 2 ã
t, t,
2t
√
(up to permutations) as long as 3 ≤ t ≤ 1. In addition, the minimum of r is equal to 0 whenever
3
t ≥ 1.
√
Thus, if 3 ≤ t ≤ 1, then
3
kt − 1
Å ã 2
2
h(t) := k − kt − 2t ≤ F(k).
2
t + 1
′
′
Note that h(t) is the same as g(t) above, so h (t) = g (t) ≤ 0, which means that
(k − 1) 2
min h(t) = h(1) = − .
h √
t∈ 3 ,1 i 2
3
Now, for z = 0, we have that

(kx − 1)(ky − 1)
F(k) = −
x + y

as xy + yz + zx = 1 turns into xy = 1.

We will show that
(kx − 1)(ky − 1) (k − 1) 2
− ≥ − ,
x + y 2
that is,
(kx − 1)(ky − 1) (k − 1) 2
≤ .
x + y 2

By cross-multiplying and taking into account the fact that xy = 1, we are left to prove the
following inequality in x and k :
1 k
Å ã Å ã
2
x + (k − 1) − 2(kx − 1) − 1 ≥ 0,
x x
which is true as it can be rearranged into

2
(k + 1)(x − 1) 2
≥ 0.
x
In conclusion,
(k − 1) 2
F(k) ≥ −
2