An inequality with cubic roots

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Proof. To prove the second inequality, we note that, since x > 2, we have. √ x + 2 > 2, ... By the AM-GM inequality we have x ≥ 2 with equality if and only if u = v.
AN INEQUALITY WITH CUBIC ROOTS SOLUTION TO PROBLEM 1822 OF THE MATHEMATICS MAGAZINE

OMRAN KOUBA Abstract. We prove the following inequality: s r  r    u v 1 1 2uv 1 3 3 ≤ (u + v) 17 − 2 + ≤ + . 8 u + v2 v u u v

Problem 1822. [1]. Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi, Vietnam. Let u and v be positive real numbers. Prove that s r   r   2uv u v 1 1 1 3 3 17 − 2 ≤ + ≤ (u + v) + . 8 u + v2 v u u v Find conditions under which equality holds. Solution [2]: We will use the following Lemma. Lemma. Let x be a real number such that x > 2. Then   √ 1 2 17 − < x < (x − 1) x + 2. 8 x(x2 − 3) Proof. To prove the second inequality, we note that, since x > 2, we have and consequently √ x < x + x − 2 = 2(x − 1) < (x − 1) x + 2. On the other hand,

√

x + 2 > 2,

2 x3 − 3x + 2 (x − 2)(x + 1)2 8x − 17 + = 8(x − 2) − = 8(x − 2) − . x(x2 − 3) x(x2 − 3) x(x2 − 3) Therefore

  2 (x + 1)2 8x − 17 + = (x − 2) 8 − . x(x2 − 3) x(x2 − 3) Writing (x + 1)2 = (x2 − 3) + 2x + 4, and noting that x2 − 3 > 1 for x > 2 we obtain (x + 1)2 1 2x + 4 1 2x + 4 5 5 9 = + < + < 2 + < 2 + = . x(x2 − 3) x x(x2 − 3) x x x 2 2

1

2

OMRAN KOUBA

Hence

  9 2 > (x − 2) 8 − > 0. 8x − 17 + x(x2 − 3) 2 The Lemma is proved.



Now r we come rto the proposed problem. If u and v are positive real numbers, we consider u v x = 3 + 3 . By the AM-GM inequality we have x ≥ 2 with equality if and only if v u u = v. So, if u 6= v we have x > 2 and by the Lemma we obtain   √ 2 1 x+2 17 − < x < (x − 1) 8 x(x2 − 3) This is equivalent to s r   r   1 2uv u v 1 1 3 3 17 − 2 < + < (u + v) + , 8 u + v2 v u u v because from x3 =

u v + + 3x, v u

we conclude that 



u v + = x3 − 3x + 2 = (x − 1)2 (x + 2) v u 2uv 2 2 = u v = 3 . 2 2 +u u +v x − 3x v Moreover, if u = v, we have equality sign in both inequalities, so this characterizes the cases of equality.  (u + v)

1 1 + u v

=2+

References [1] Pham Van Thuan, Proposed Problem 1822, Mathematics Magazine, Vol. 82, No. 3, (2009), p. 227. www.jstor.org/stable/27765909 [2] Omran Kouba, An invariant Ratio: 1809, Mathematics Magazine, Vol. 83, No. 3, (2010), pp. 229– 230. www.jstor.org/stable/10.4169/mathmaga.83.3.227a Department of Mathematics, Higher Institute for Applied Sciences and Technology, P.O. Box 31983, Damascus, Syria. E-mail address: omran [email protected]