Introduction

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In[7],theauthorusedanelementarymethodgaveashortproofofthewell-knownHoldersinequality:

whereallak bk>0and =1withp q>1 Theequalityholdswhen

forallk j=1 2 nMoreover,ifp=q=2,inequality(1)reducestothewell-knownCauchysinequality:

In[4,8-10], theauthorsconsideredthefollowingfunction

andprovedthat for0 t₁ < t₂ << t_k 1 thenthe following interpolationandre nementof the Holdersinequality

Like Holders in equality, the well-known Minkowskis in equality plays also an important rule in themath ematical and physics research elds and literatures. There are the most important, interesting,useful and elementary in equalities in mathematics,physics and other research elds. It plays an important role in math ematics and physics research elds and has great potential in the future research.There were many research papers devoted to the generalizations,re nements and applications of these two important inequalities.For examples, were fert other references in[1-11]and there referencescited in them. Since their importance and applicationpotential both in theory and practical applications, in this paper, inspired by the works of [4,7 10],weare going to present a short proof of the well-known Minkowskis inequality and give are nement andan inter polation of it.Ourresults are new and givenb elow.

Theorem 1. If a_i b_i>0 i=12 np>0, thenforp 1,wehave

and for 0< p 1,we have

Theorem2. DeneaC functiong(x)asfollows

Then for p=1,g(x) (ak+bk) g(0)=constant.

For p>1 g(x) 0 and for0 x1< x2< < xm 1,the following inequalities are re nementsand interpolation of g(x):

For 0< p< 1g(x) 0 and for0 x1< x2< < xm 1, the following in equalities are re nements and inter polations of g(x).

Moreover,

and g (x) 0 if and only if g(1)=g(0), in this case,g(x) 0 and g(x) g(0)=constant

Proof of Theorem1. For1<=m<=n set x=a _m and define

where

Hence

Solvingequationf(x)=0,weget

and after some calculations,we obtain

It follows from equation(9)and x= a_m that

That is constant Letm=1 2 n,we get b_k = a_k k=12 n. Substituting above equations into(6), we get

Since

wegetfrom(8) (9)that

If p=1, thenfm(x) 0, inthiscase, (4)and(5)becomes equality for all a_k>0 b_k>0 1< k< n.

If p>1, then fm(x)>0 for all x>0,hence f_m(x) achieves its minimum at x=x where x satisfies f_m(x)=0 in this case in equality(4) holds.

If 0< p< 1,then f_m(x)< 0 for all x> 0,hence f_m(x)achieves its maximum at x=x where x satisfies f_m(x)=0 in this case, inequality (5) holds.Theorem 1 is proved.

Proof of Theorem 2. It follows from the expression of g(x) that

By Theorem 1, if p=1, then

If p > 1, then g(1) g(0),with equality holds if and only if Inthis case,

Then (7) is an interpolation and a re nement of (4).

If 0 < p < 1, then g(1) g(0), with equality holds if and only if in this case,

Declarations

Ethical Approval

Consent to Publish.

Funding

Not Applicable

Data availability

No data was used for the research described in the article.

Conflict of Interest

Not Applicable

References

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