On the Cîrtoaje’s conjecture

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R E S E A R C H

Open Access

On the Cîrtoaje’s conjecture

Ladislav Matejíˇcka

*

*_{Correspondence:}

ladislav.matejicka@tnuni.sk Faculty of Industrial Technologies in Púchov, Trenˇcín University of Alexander Dubˇcek in Trenˇcín, I. Krasku 491/30, Púchov, 02001, Slovakia

Abstract

In this paper, we prove the Cîrtoaje conjecture under other conditions.

MSC: Primary 26D10; secondary 26D15

Keywords: inequality; power inequalities; exponential inequalities; power exponential functions

1 Introduction and preliminaries

Within the past years, the power exponential functions have been the subject of very in-tensive research. Many problems concerning inequalities for the power exponential func-tions look so simple, but their solufunc-tions are not as simple as it seems. A lot of interesting results for inequalities with the power exponential functions have been obtained. The his-tory and a literature review of inequalities with power exponential functions can be found for example in []. Some other interesting problems concerning stronger inequalities of power exponential functions can be found in []. In the paper, we study one inequality conjectured by Cîrtoaje []. Cîrtoaje, in [], has posted the following conjecture on the inequalities with power exponential functions.

Conjecture . If a,b∈(; ]and r∈[;e],then

√ara_brb≥_arb₊_bra_. _(.)

The conjecture was proved by Matejíčka []. In [], Coronel and Huancas posted several conjectures for inequalities with the power exponential functions. Some of them are not valid as was shown by the unknown referee of this paper. We note that Theorems ., ., Lemma ., Conjectures . and . in [] are not valid. With the referee’s kind permission we present his counterexample:n= ,x=_,x=_,x=_,r=_, when

xrx

 +x

rx

 +x

rx

 <x

rx

 +x

rx

 +x

rx

 .

In this paper, we show that the conjecture (.) is also valid under the conditions: ≤

min{a,b} ≤max{a,b} or  <min{a,b} ≤/e and max{a,b} ≥, or  <min{a,b} ≤e,

max{a,b} ≥e, andr∈[;e].

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2 Main results

Theorem . Let a,b be positive numbers.The inequality

√ara_brb_≥_arb₊_bra _(.)

holds for any r∈ ,eif one of the following three conditions is satisﬁed:

(a) a≥b≥;

(b) a≥≥ e

b;

(c) a≥e≥b.

Proof According to the power mean inequality, it suﬃces to consider the case wherer

has the maximum value, that isr=e. Without loss of generality, supposea≥band de-note

H(x) = xex_beb_–_xeb_–_bex_.

An easy calculation gives

H(x) =exex_beb₍_ln_x_{+ ) –}_bxeb–_–_bex_ln_b_,

H(x) =e

xex_beb

e(lnx+ )

 +



x

–b(eb– )xeb––ebexlnb

.

Solution for (a):a≥b_≥_.

Supposex≥b≥. We showH(x)≥,H(b)≥, andH(b)≥. It impliesH(a)≥ fora≥b_≥_.

It is easy to see that _eH(x)≥U+V, where

U=xex_beb_e_ln_b_–_ebex_ln_b_,

V=xex_beb

elnb

 +

e



+xex–_beb _–_eb_xeb–₊_bxeb–_.

Similarly, we estimate

U≥ebex_ln_b_beb _–_ln_b≥_ebex_ln_b_.

It follows from

f(b) = eb

 lnb–ln( +lnb)≥,

because of

f(b) =eb( +lnb)

_{– }

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Next we estimate

V=xeb–

xex–eb+_beb

elnb

 +

e



+xex+–eb_beb _–_eb₊_b

≥xeb–

beb–eb++eb

elnb

 +

e



+beb+–eb+eb _–_eb₊_b

.

Using

l(b) =eb– eb+  +eb

 ≥

 –e

 , m(b) =eb

_{– }_eb_{+  +}eb

 ≥

 –e

 ,

l(b) = eb– e+e

 > , m

₍_b_{) = }_eb_{– }_e₊e

 > , l() =  –e

 ,

m() =  –e

 ,

 –e

 >  ,

 –e

 >  ,

we haveH(x)≥ forx≥b_≥_{ if}

p(b) = e b



 _–_eb₊_b₊_b ≥_.

Putu=b/then we obtainp(b)≥ if

k(u) = e u

_–_eu₊_u_{+ }_≥_.

It is easy to see thatk(u) =  if

q(u) = eu– eu+  = .

From Cardano’s formula we see that the root of q(u) is equal to u = .. From

k(.) = . and from k(u) = eu – eu≥ we havek(u)≥ foru≥. So

H(x)≥ forx≥b_≥_.

Now we showH(b₎_≥_.

Some calculation gives

Hb=ebeb+eb_(_ln_b_{+ ) –}_beb–_–_beb_ln_b_.

From this we haveH(b₎_≥_{ if}

beb+eb _–_beb_ln_b₊_beb+eb₍_ln_b_{+ ) –}_beb–≥_.

Rewriting this we obtainH(b₎_≥_{ if}

beb–beb+eb–eb+₍_ln_b_{+ ) – }₊_beb_ln_b_beb _{– }≥_.

But this will be fulﬁlled if

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This is equivalent to

o(b) =

eb– eb

 + 

lnb+ln(lnb+ )≥.

Usinglnb≥b–_b we haveo(b)≥ if we show that

s(b) =

eb– eb 

lnb+ln(b– )≥.

s(b) =

eb– e 

lnb+eb– e

+

 b– .

Because ofs() =  it suﬃces to show thateb– _e +__b_–≥. But this is evident from  – e

≥ andeb– 

e

+  b–=



b–(e(b– ) _{+  –}e

).

Now we show thatH(b₎_≥_.

Hb= beb_beb _–_beb_–_beb₌_beb__beb+eb–eb_–_beb–eb_{– }≥_

if

beb+eb–eb_–_beb–eb₌_beb–eb__beb _{– }≥_.

It suﬃces to show that beb _{– }≥_be_{because of}

beb–ebbeb _{– }≥_be(b–)≥_.

DenoteF(b) = beb –be. Then

F(b) =ebeb₍_ln_b_{+ ) –}_be–_.

Because ofF() =  it suﬃces to show thatF(b)≥. We will be done if we show that

s(b) =

eb

 –e

lnb+ln(b– )≥.

We usedlnb≥(b– )/b. Because ofs() =  it suﬃces to show that

s(b) = e lnb+

eb₊_b_{( – }_e_{) + }_e

b(b– ) ≥.

But this is evident because ofj(b) = eb₊_b_{( – }_e_{) + }_e_≥_{. (}_j_{() =  –}_e_≥_,_j_{() =  –}_e_≥

,j(b) = e≥.)

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Ifb≤/eandx≥ then it suﬃces to show that

H(x) =exex_beb₍_ln_x_{+ ) –}_bxeb–_–_bex_ln_b≥_

andH()≥.

It is evident thatH(x)≥ if

xex–eb+_beb–₍_ln_x_{+ )}≥_.

It follows from ex

 –eb+ ≥ and from

eb

 – ≤. Finally, we show thatH()≥. Denote

v(b) = beb _{–  –}_be_.

Then we havev(/e) = /e–  – (/e)e_{= .. If we show}_v₍_b₎_≤_{ then the conjecture}

will be proved. An easy calculation gives

v(b) =ebeb₍_ln_b_{+ ) –}_be–_.

v(b)≤ if

beb–e+₍_ln_b_{+ )}≤_.

If ≤b≤/ethen the inequality is evident. Let /e≤b≤/e, thenv(b)≤ if

s(b) =

eb

 –e+ 

lnb+ln(lnb+ )≤.

Numerical calculation shows thats(/e) = –.. So it suﬃces to show that

s(b) = e lnb+

eb

 –e+ 



b+



lnb+  

b≥.

This is equivalent to



(lnb+ ) + 

 +lnb +  –e≥. (.)

(We usedeb/≥/.)

Putt=  +lnb. Then (.) can be rewritten as



tm(t) =



t

t+ ( –e)t+ ≥

for  <t≤ln. Fromm(t) = t+ ( –e)≤–. <  and fromm(ln) =ln +  + ( –

e)ln = .≥ we see that the proof of the case (b) is complete. Solution for (c)a≥e≥b.

It suﬃces to show that√ara_brb_≥_arb_and√_ara_brb_≥_bra_{. The ﬁrst inequality is equivalent}

to

aa–b≥

a b

b

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or

ax–≥x, x=a

b≥.

Indeed, we have

ax–≥ex–≥x.

The second inequality is equivalent to

b a a

ba–b≤, or

xb–x≤, x=b

a≤.

Indeed we have

xb–x≤xe–x≤.

This completes our proof.

Note . We note that the proof of the case (c) originates from an unknown reviewer. His proof of (c) is more elegant and his formulation of (c) is more general than ours.

Note . We note that

√e_e_{– }e_{– }e_{= –.}_e_,

which implies that the inequality (.) is not valid for alla,b≥.

The referee’s counterexample implies that (.) cannot be generalized forn= .

Competing interests

The author declares that he has no competing interests.

Acknowledgements

The work was supported by VEGA grant No. 1/0385/14. The author thanks the faculty FPT TnUAD in Púchov, Slovakia for its kind support and he is deeply grateful to the unknown reviewer for his valuable remarks, suggestions, and some generalization in the Theorem 2.1, for his kind permission to publish his counterexample, his version of the proof of (c) in Theorem 2.1.

Received: 31 August 2015 Accepted: 21 May 2016 References

1. Coronel, A, Huancas, F: The proof of three power exponential inequalities. J. Inequal. Appl.2014, 509 (2014) http://www.journaloﬁnequalitiesandapplications.com/content/2014/1/509

2. Miyagi, M, Nishizawa, Y: A stronger inequality of Cîrtoaje’s one with power-exponential functions. J. Nonlinear Sci. Appl.8, 224-230 (2015) http://www.tjnsa.com

3. Cîrtoaje, V: Proofs of three open inequalities with power exponential functions. J. Nonlinear Sci. Appl.4(2), 130-137 (2011)