where, g(x) = 2x3 + 6x2 sin x cosx − 5 sin3 x cosx − 3x sin2 x − 2xsin4 x. Direct ... x − x3. 3! + x5. 5! ) − 9 sin2 x cos x − 4x sin3 x. = 9x2 −. 7. 2 x4 +. 11. 40.
M athematical I nequalities & A pplications Volume 11, Number 1 (2007), 149–151
A NEW ELEMENTARY PROOF OF WILKER’S INEQUALITIES LU ZHANG AND LING ZHU (communicated by P. Bullen)
Abstract. In this note, we show a new elementary proof of Wilker’s inequalities.
1. Introduction J. B. Wilker [1] proposed two open questions and the second one was the following statement: (A) There exists a largest constant c such that 2 sin x tan x > 2 + cx3 tan x + (1) x x for 0 < x < π2 . J. S. Sumner et al. [2] affirmed the truth of the Problem (A) and obtained a further results as follows THEOREM. If 0 < x < π /2 , then 2 8 3 sin x tan x 16 3 −2< x tan x. x tan x < + π4 x x 45
(2)
Furthermore, 16/π 4 and 8/45 are the best constants in (2). B. N. Guo et al. [3] gave a new proof of the inequalities (2) by using the power series expansion of cotangent function and a lower bound of Bernoulli numbers. Recently, I. Pinelis [4] got other proof of inequality (2) by using L’Hospital rules for monotonicity. In this note, we show a new elementary proof of Wilker’s inequalities. 2. One Lemma LEMMA. If x > 0 , then x2 x2 x4 < cos x < 1 − + , 2! 2! 4! x3 x5 x7 x3 x5 x− + − < sin x < x − + . 3! 5! 7! 3! 5! 1−
Key words and phrases: Wilker’s inequalities. c , Zagreb Paper MIA-11-09
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LU ZHANG AND LING ZHU
3. A New Elementary Proof of Theorem Let f (x) =
1 2 sin2 x + , then − 3 x5 tan x x4 x tan x f (x) =
g(x) , x6 sin2 x
where, g(x) = 2x3 + 6x2 sin x cos x − 5 sin3 x cos x − 3x sin2 x − 2x sin4 x . Direct calculation yields g (x) = 2 cos xh(x), where, h(x) = 6x2 cos x + 3x sin x − 9 sin2 x cos x − 4x sin3 x. Then applying the inequalities in Lemma, we obtain x4 x5 x2 x3 2 + + h(x) < 6x 1 − + 3x x − − 9 sin2 x cos x − 4x sin3 x 2! 4! 3! 5! 7 11 = 9x2 − x4 + x6 − 9 sin2 x cos x − 4x sin3 x = q(x), 2 40 and 33 5 x − 18 sin x cos2 x + 5 sin3 x − 12x sin2 x cos x 20 33 = 18x − 14x3 + x5 − 18 sin x + 23 sin3 x − 12x sin2 x cos x 20 x5 x7 x5 33 x3 x3 < 18x − 14x3 + x5 − 18 x − + − + + 23 sin2 x x − 20 3! 5! 7! 3! 5! 2 x − 12x sin2 x 1 − 2! 3 1 7 13 23 5 x + 11x + x3 + x sin2 x = −11x3 + x5 + 2 280 6 120 x5 3 5 1 7 13 3 23 5 x3 3 x + 11x + x + x + x− < −11x + x + sin x 2 280 6 120 3! 5! 3 5 1 7 1 4 7 6 1 8 23 10 3 2 x + 11x + x − x − x + x sin x = −11x + x + 2 280 3 90 72 14400 3 1 7 x < −11x3 + x5 + 2 280 x5 1 7 1 23 10 x3 x + + 11x2 + x4 − x6 − x8 + x− 3 90 72 14400 3! 5! 1 7 1 9 17 11 1 4 5 x + x + x − x + x13 = x2 − 105 540 300 43200 72000
