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40 Dorin MARGHIDANU

so the first two inequalities in the statement. For the third inequality, we note
2
that: (1 + px) [1 + (r − p)x] = 1 + rx + p(r − p)x ≥ 1 + rx.
b) Similar proof, but with the application of inequality (B) for the powers r −p
and p, successively.
If we decompose the initial power into product by three factors, we can extend
the refinement:
Proposition 5 (Triple refinement of Bernoulli’s inequality). If x ≥ 0 and
m, n, p, r ∈ R ≥1 such that r = m + n + p, then

r
p
n
(1 + x) ≥ (1 + mx)(1 + x) (1 + x) ≥ (1 + mx)(1 + nx)(1 + x) p
≥ (1 + mx)(1 + nx)(1 + px) ≥ 1 + rx. (B7)
Proof. We apply inequality (B) for the powers m, n and p, successively:

(B)
r
p
m
n
n
(1 + x) = (1 + x) (1 + x) (1 + x) ≥ (1 + mx)(1 + x) (1 + x) p
(B) (B)
p
≥ (1 + mx)(1 + nx)(1 + x) ≥ (1 + mx)(1 + nx)(1 + px),
that is the first three inequalities in the statement. The fourth inequality results
as follows:
2
(1 + mx)(1 + nx)(1 + px) = 1 + (m + p + n)x + (mn + np + pm)x + mnpx 3
2
3
= 1 + rx + (mn + np + pm)x + mnpx ≥ 1 + rx.
Remark 2. Similar to refinement (B7), five more triple-refinements can be obtained
by permutations, applying the inequality (B) successively for the powers: (m, p, n),
(n, m, p), (n, p, m), (p, m, n), (p, n, m). Obviously, according to the above models,
multiple refinements of Bernoulli inequality can be obtained. The refinements can
also be adapted for the case of r - subunit powers.

References

[1] J. Bernoulli, Positiones arithmeticae de seriebus infinitis earumque summa finita,
(Treatise on Infinite Series), Basel, 1689.
[2] P.S. Bullen, Equivalent Inequalities, arXiv 0809.0641 v2 [math.CA], 2008.

[3] D.S. Mitrinovi´c, J.E. Peˇcari´c, Bernoulli’s Inequality, Rendiconti del Circolo Matematico
di Palermo, vol. 42, no. 3, 1993, pp. 317-337.

[4] D.S. Mitrinovi´c, J.E. Peˇcari´c, A.M. Fink, Classical and New Inequalities in Analysis,
Kluwer Acad. Press., 1993, pp. 65-82.