**R E S E A R C H**

**Open Access**

## Improved estimates for the triangle

## inequality

### Nicu¸sor Minculete

*_{and Radu P˘alt˘anea}

*_{Correspondence:}
minculeten@yahoo.com
Faculty of Mathematics and
Computer Science, Transilvania
University of Bra¸sov, Str. Iuliu Maniu,
nr. 50, Bra¸sov, 500091, Romania

**Abstract**

We obtain reﬁned estimates of the triangle inequality in a normed space using integrals and the Tapia semi-product. The particular case of an inner product space is discussed in more detail.

**MSC:** Primary 46B99; secondary 26D15; 46C50; 46C05

**Keywords:** norm inequalities; triangle inequality; Tapia semi-product; inner product
spaces; norm-angular distance

**1 Introduction**

The theory of inequalities plays an important role in many areas of Mathematics. Among the most used inequalities we ﬁnd the triangle inequality. This inequality is the following:

*x*+*y* ≤ *x*+*y*,

for any vectors*x* and*y*in the normed linear space (*X*, · ) over the real numbers or
complex numbers. Its continuous version is,

* _{a}bf*(

*t*)

*dt*≤

*b*
*a*

*f*(*t*)*dt*,

where*f* : [*a*,*b*]⊂R→*X*is a strongly measurable function on the compact interval [*a*,*b*]
with values in a Banach space*X*and*f*(·)is the Lebesgue integrable on [*a*,*b*].

Diaz and Metcalf [] proved a reverse of the triangle inequality in the particular case of
spaces with inner product. Several other reverses of the triangle inequality were obtained
by Dragomir in []. Also, in [] some inequalities for the continuous version of the triangle
inequality using the Bochner integrable functions are given. In [], Rajić gives a
charac-terization of the norm triangle equality in pre-Hilbert*C*_{-modules. In [, ], Maligranda}

proved a reﬁnement of the triangle inequality as follows.

**Theorem A** *For nonzero vectors x and y in a normed space*(*X*, · )*it is true that*

– *x*
*x*+

*y*
*y*

min*x*,*y*≤ *x*+*y*–*x*+*y*

≤

– *x*
*x*+

*y*
*y*

max*x*,*y*. ()

In [] Kato, Saito and Tamura proved the sharp triangle inequality and reverse inequality
in Banach space for nonzero elements*x*,*x*, . . . ,*xn*∈*X*, which is in fact a generalization

of Maligranda’s inequality. Another extension of Maligranda’s inequality for*n*elements in
a Banach space was obtained in Mitani and Saito []. The problem of characterization of
all intermediate values*C*satisfying ≤*C*≤*n _{k}*

_{=}

*xk*–

*n*

*k*=*xk*, for*x*,*x*, . . . ,*xn*in a

Banach space is studied by Mineno, Nakamura and Ohwada [], Dadipour*et al.*[], Sano
*et al.*[] and others. For other diﬀerent results about the triangle inequality we mention
only [–].

The main aim of the present paper is to provide an improvement of the inequality due
to Maligranda. Some other estimates which follow from the triangle inequality are also
presented. Moreover, we can rewrite them as estimates for the so-called*norm-angular*
*distance*or*Clarkson distance*(see*e.g.*[]) between nonzero*x*and*y*as*α*[*x*,*y*] =_{}*x _{x}*

_{}–

_{}

*y*

_{y}_{}. This distance was generalized to the

*p*-angular distance in normed space in []. In [], Dragomir characterizes this distance obtaining new bounds for it. A survey on the results for bounds for the angular distance, so named Dunkl-Williams type theorems (see []), is given by Moslehian

*et al.*[].

**2 Estimates of the triangle inequality using integrals**

Let (*X*, · ) be a real normed space. The following lemma is evident.

**Lemma ** *For any x*,*y*∈*X*,*the function g*(*s*) =*x*+*sy*,*s*∈R,*is convex*.

**Theorem ** *For any x*,*y*∈*X we have*

(i) *x*+*y* ≤ _{}( –*λ*)*x*+*λydλ*≤ *x*+*y*,
(ii) *x*+*y*+*x*+*y* ≥ _{}( –*λ*)*x*+*λydλ*.

*Proof* (i) Let*a*,*b*∈R,*a*<*b*arbitrarily chosen. Since the function*g*(*s*) =*z*+*su*,*s*∈Ris
convex, for any ﬁxed*u*,*z*∈*X*we apply the Hermite-Hadamard inequality, see*e.g.*[, ]
for*a*<*b*:

*g*

*a*+*b*

≤

*b*–*a*

*b*
*a*

*g*(*s*)*ds*≤*g*(*a*) +*g*(*b*)

.

Therefore we obtain

*z*+*a*+*b*

*u*

≤

*b*–*a*

*b*
*a*

*z*+*suds*≤*z*+*au*+*z*+*bu*

. ()

For any*a*<*b*and any*x*,*y*∈*X*,*u*=_{a}_{–}* _{b}*(

*x*–

*y*) and

*z*=

_{a}_{–}

*(*

_{b}*ay*–

*bx*) such that

*x*=

*z*+

*au*and

*y*=

*z*+

*bu*. By replacing in inequality () we deduce the inequality

*x*+*y* ≤
(*b*–*a*)

*b*
*a*

*bx*–*ay*+*s*(*y*–*x*)*ds*≤*x*+*y*

.

If we multiply these inequalities by and make the change of variable*s*= ( –*λ*)*a*+*λb*in
the integral, we arrive at relation (i).

(ii) If we apply the Hammer-Bullen inequality, see*e.g.*[], for*a*<*b*we obtain

*g*(*a*) +*g*(*b*)

+*g*

*a*+*b*

≥

*b*–*a*

*b*
*a*

Now, if we proceed similarly to (i) we obtain relation (ii).

**Example ** Let*X*=R_{, endowed with the norm}_{}_{(}_{x}

,*x*)=max{|*x*|,|*x*|}. Let*η*,*μ*∈[, ]

and deﬁne*x*= (, ),*y*= (–*μ*,*ημ*). We have*x*= ,*y*=*μ*,*x*+*y*= +*ημ*. For*λ*∈[, ],
since ( –*λ*)(–*λη*)≤ and ( –*λ*)*λημ*≥, we obtain( –*λ*)*x*+*λy*=( –*λ*–*λμ*, –*λ*+
*λημ*)= –*λ*+*λημ*and then

( –*λ*)*x*+*λydλ*=

( –*λ*+*λημ*)*dλ*= +*ημ*.

Hence, relations (i) and (ii) become

(i) +*ημ*≤ +*ημ*≤ +*μ*, (ii) +*μ*+*ημ*≥ + *ημ*.

**Remark ** Since the parameters*η*,*μ*can be taken arbitrarily in the interval [, ] in
Ex-ample , it follows that the constants in front of the terms of inequalities (i) and (ii) in
Theorem are optimal.

**Corollary ** *For nonzero elements x*,*y from a space with inner product X*= (*X*,·,· )*and*
*any a*,*b*∈R,*a*<*b*,*we have*

(*xy*–*x*,*y*)

*x*+*y*+ _{}( –*λ*)*x*+*λydλ*≤ *x*+*y*–*x*+*y*

≤ (*xy*–*x*,*y* )

*x*+*y*+ _{}( –*λ*)*x*+*λydλ*. ()

*Proof* We have*x*+*y*_{=}_{}_{x}_{+}_{y}_{,}_{x}_{+}_{y}_{=}_{}_{x}_{}_{+ }_{x}_{,}_{y}_{+}_{}_{y}_{}_{, which implies}

*x*+*y*=*x*+*y*– *xy*–*x*,*y*,

which means that
(*xy*–*x*,*y* )

*x*+*y*+*x*+*y*=*x*+*y*–*x*+*y*.

Using point (i) from Theorem in the above equality we obtain the inequalities of the

statement.

**Remark ** Inequality () represents an improvement of the Cauchy-Schwarz inequality.

**3 Estimates of the triangle inequality using the Tapia semi-product**

The Tapia semi-product on the normed space*X*(see []) is the function (·,·)*T*:*X*×*X*→
R, deﬁned by

(*x*,*y*)*T*=lim
*t*

*ϕ*(*x*+*ty*) –*ϕ*(*x*)

*t* ,

|(*x*,*y*)*T*| ≤ *xy*for all*x*,*y*∈*X*. In the case when the norm · is generated by an inner

product·,· , then (*x*,*y*)*T*=*x*,*y*, for all*x*,*y*∈*X*.

For instance, inR*n*_{, with}_{p}_{-norm, <}_{p}_{<}_{∞}_{,}_{}_{x}_{}_{= (}*n*

*i*=|*xi*|*p*)/*p*, for*x*= (*x*,*x*, . . . ,*xn*),

the Tapia semi-product becomes

(*x*,*y*)*T*=

*n*

*i*=*αiyi*|*xi*|*p*–

(*n _{i}*

_{=}|

*xi*|

*p*)

*p*–

*p*
,

if*x*= , where

*αi*=

, if*xi*= ,
*xi*

|*xi*|, otherwise.

For nonzero elements*x*,*y*∈*X*denote

*v*(*x*,*y*) = *x*
*x*+

*y*

*y*. ()

First from the Maligranda result we deduce the following inequality.

**Theorem ** *Let x*,*y*∈*X*,*be nonzero vectors*.*Then we have*

(*x*,*y*)*T*≤ *x* · *yv*(*x*,*y*)–

. ()

*Proof* If in the left inequality of Theorem A we replace*y*by*ty*, with*t*> ,*t*<*x*/*y*and
then we divide them by*t*we obtain

–*v*(*x*,*y*)*y* ≤ *y*–*x*+*yt*–*x*

*t* .

Letting*t*→ we obtain

–*v*(*x*,*y*)*y* ≤ *y*–(*x*,*y*)*T*
*x* .

From this we obtain immediately equation ().

**Remark ** Inequality () is an improvement of the known inequality (*x*,*y*)*T*≤ *x* · *y*,

since*v*(*x*,*y*) ≤.

**Theorem ** *Let x*,*y*∈*X*,*be nonzero vectors such thaty* ≤ *xandxy*= –*yx*.*Then*

*x*+*y*–*x*+*y* ≥ –*v*(*x*,*y*)*x*
–

–

*x*+*y*
*x*+*y*,

*y*
*y*

*T*

*x*–*y*, ()

*x*+*y*–*x*+*y* ≤ –*v*(*x*,*y*)*x*
–

–

*v*(*x*,*y*)
*v*(*x*,*y*),

*y*
*y*

*T*

*x*+*y*–*x*+*y* ≥ –*v*(*x*,*y*)*y*
+

–

*x*+*y*
*x*+*y*,

*x*
*x*

*T*

*x*–*y*, ()

*x*+*y*–*x*+*y* ≤ –*v*(*x*,*y*)*y*
+

–

*v*(*x*,*y*)
*v*(*x*,*y*),

*x*
*x*

*T*

*x*–*y*. ()

*Proof* We choose the following notations:

*z*= *x*

*x*, *u*=

*y*

*y*, *λ*=

*y*

*x*, *μ*=

*x*

*y*. ()

For obtaining the ﬁrst two inequalities we consider the function*f* : [,∞)→R,

*f*(*s*) = +*s*–*z*+*su*, *s*∈[,∞).

Also denote

*g*(*s*) =*z*+*su*, *s*∈[,∞).

The conditions given in the theorem ensure that*z*+*su*= , for*s*∈[,∞). We have

*z*+*su*
*z*+*su*,*u*

*T*

=

*z*+*su*(*z*+*su*,*u*)*T*

=lim
*t*

*z*+ (*s*+*t*)*u*_{–}_{}_{z}_{+}_{su}_{}

*tz*+*su*

=lim
*t*

*z*+ (*s*+*t*)*u*–*z*+*su*
*t*

=lim
*t*

*g*(*s*+*t*) –*g*(*s*)
*t*

=*g*(*s*+ ),

where we denoted by*g*(*s*+ ) the right derivative of*g*at point*s*. Then

*f*(*s*+ ) = –

*z*+*su*
*z*+*su*,*u*

*T*

, *s*∈[,∞). ()

We show that for*λ*∈[, ] we have

*f*() – ( –*λ*)*f*(*λ*+ )≤*f*(*λ*)≤*f*() – ( –*λ*)*f*( + ). ()

Because these inequalities are obvious for*λ*= , we suppose that ≤*λ*< .

Choose an arbitrary number*t*> . The function*f* is the diﬀerence between a linear and
a convex function. Then it is concave. Consequently we have

*t*–
*t*–*λf*(*λ*) +

–*λ*

Straightforward computations shows that this inequality can be written equivalently, in the form

*f*(*λ*)≤*f*() – ( –*λ*)*f*(*t*) –*f*()
*t*– .

If we pass to the limit*t*→,*t*> in this inequality we obtain the right side inequality in
().

Also, choose an arbitrary number*λ*<*t*< . Since*f* is concave we have

*t*–*λ*
–*λf*() +

–*t*

–*λf*(*λ*)≤*f*(*t*).

After short computations, this inequality can be written equivalently, in the form

*f*() – ( –*λ*)*f*(*t*) –*f*(*λ*)
*t*–*λ* ≤*f*(*λ*).

Passing to the limit*t*→*λ*,*t*>*λ*we obtain the left side inequality in ().
From equation () we obtain

*f*(*λ*+ ) = –

*x*+*y*
*x*+*y*,

*y*
*y*

*T*

, *f*( + ) = –

*v*(*x*,*y*)
*v*(*x*,*y*),

*y*
*y*

*T*

. ()

If we multiply equations () by*x*and we take into account relations () and also the
following relations:

*xf*(*λ*) =*x*+*y*–*x*+*y*,
*xf*() =*x* –*v*(*x*,*y*),
*x*( –*λ*) =*x*–*y*,

we deduce inequalities () and ().

In order to obtain the last two inequalities we consider the function*f*: [,∞)→R,

*f*(*s*) = +*s*–*sz*+*u*, *s*∈[,∞).

The right derivatives of the function*f*can be obtained immediately from the derivatives

of the function*f* by interchanging*u*and*z*, see equation (). So we obtain

*f*_{}(*s*+ ) = –

*sz*+*u*
*sz*+*u*,*z*

*T*

, *s*∈[,∞). ()

For*μ*≥ we show the double inequality:

*f*() +*f*(*μ*+ )(*μ*– )≤*f*(*μ*)≤*f*() +*f*( + )(*μ*– ). ()

We consider only the case*μ*> , since for*μ*= these inequalities are obvious.
First take an arbitrary number*t*>*μ*. Since*f*is concave, we have

*t*–*μ*

*t*– *f*() +
*μ*–

This inequality can be rewritten in the equivalent form

*f*() +

*f*(*t*) –*f*(*μ*)

*t*–*μ* (*μ*– )≤*f*(*μ*).

If we pass to the limit*t*→*μ*,*t*>*μ*we obtain the left side inequality in ().
Next let us take an arbitrary number <*t*<*μ*. Since*f*is concave we have

*t*–
*μ*– *f*(*μ*) +

*μ*–*t*

*μ*– *f*()≤*f*(*t*).

We can transform this inequality to the form

*f*(*μ*)≤*f*() +

*f*(*t*) –*f*()

*t*– (*μ*– ).

If we pass to the limit*t*→,*t*> we obtain the right side inequality in (). If we multiply
inequalities given in () by*y*and take into account the relations

*yf*(*μ*) =*x*+*y*–*x*+*y*,

*yf*() =*y*

–*v*(*x*,*y*),
*f*_{}( + ) = –

*v*(*x*,*y*)
*v*(*x*,*y*),

*x*
*x*

*T*

,

*f*_{}(*μ*+ ) = –

*x*+*y*
*x*+*y*,

*x*
*x*

*T*

,

*y*(*μ*– ) =*x*–*y*,

we obtain relations () and ().

**Remark ** Inequalities () and () improve inequality ().

**Example ** Let the space*X*and the vectors*x*,*y*be exactly like in Example . Then we
have *x*+*y*–*x*+*y*=*μ*( –*η*). Also *v*(*x*,*y*) = (, ) + (–,*η*) = (, +*η*). Then ( –
*v*(*x*,*y*))*x*= –*η*. Hence the right inequality in () of Theorem A reads*μ*( –*η*)≤ –*η*.
We obtain an improvement of this inequality by using inequality () given in Theorem .
Indeed we have

*v*(*x*,*y*)
*v*(*x*,*y*),

*y*
*y*

*T*

=(, ), (–,*η*)* _{T}*=lim

*t*

*t*(–*t*, +*tη*)

–(, )=*η*.

It follows ( – (_{}*v _{v}*(

_{(}

*x*,

_{x}_{,}

*y*)

_{y}_{)},

_{}

*y*

_{y}_{})

*T*)(

*x*–

*y*) = ( –

*η*)( –

*μ*). Consequently inequality () from

Theorem becomes*μ*( –*η*)≤*μ*( –*η*).

It is easy to see that we can write*α*[*x*,*y*] =*v*(*x*, –*y*). Using inequalities () and () we
deduce the following double inequality.

**Corollary ** *For nonzero vectors x and y*,*such thatxy*=*yx*,

*x*–*y*–|*x*–*y*|

min{*x*,*y*} +*A*≤*α*[*x*,*y*]≤

*x*–*y*+|*x*–*y*|

*where*

*A*=

–

*x*–*y*
*x*–*y*,

*x*
*x*

*T*

_{|}

*x*–*y*|

min{*x*,*y*}≥,

*B*=

–

*v*(*x*, –*y*)
*v*(*x*, –*y*), –

*y*
*y*

*T*

_{|}

*x*–*y*|

max{*x*,*y*}≥.

*Proof* Because of the symmetry we can consider that*x* ≥ *y*.

**Remark ** Inequalities () improve the inequalities for the norm-angular distance, of
Maligranda [], which can be obtained from (). Other inequalities for the norm-angular
distance could be obtained combining all inequalities (), (), (), and ().

**4 Inequalities in inner product spaces**

In this section we derive inequalities in an inner product space (*X*,·,· ) from Theorem ,
by taking into account that (*x*,*y*)*T*=*x*,*y*,*x*,*y*∈*X*.

**Theorem ** *Let*(*X*,·,· )*be an inner product space*,*with norm* · .*For nonzero elements*
*x*,*y*∈*X*

*x*+*y*–*x*+*y* ≤

–

*v*(*x*,*y*)

*x*+*y*, ()

*where*

*v*(*x*,*y*)= *x*
*x*+

*y*
*y*
=

+ *x*,*y*
*x* · *y*

. ()

*Proof* Because of the symmetry we can suppose that*x* ≥ *y*. First we have

*v*(*x*,*y*)=

*x*
*x*+

*y*
*y*,

*x*
*x*+

*y*
*y*

=

+ *x*,*y*
*x* · *y*

.

Hence we get (). Next we obtain

*v*(*x*,*y*)
*v*(*x*,*y*),

*y*
*y*

*T*

=

*y* · *v*(*x*,*y*)
*x*
*x*+

*y*
*y*,*y*

=

*v*(*x*,*y*)

_{}

*x*,*y*
*x* · *y*+

=

*v*(*x*,*y*). ()

We can apply Theorem . From equation (), by taking into account equation () we obtain

*x*+*y*–*x*+*y* ≤ –*v*(*x*,*y*)*x*
–

–

*v*(*x*,*y*)
*v*(*x*,*y*),

*y*
*y*

*T*

= –*v*(*x*,*y*)*x*–

–

*v*(*x*,*y*)

*x*–*y*

=

–

*v*(*x*,*y*)

*x*+*y*. _{}

**Remark ** From the proof we can see that, in an inner product space, inequality () is
equivalent to inequality (). In a similar way we can obtain

–*v*(*x*,*y*)+

–

*v*(*x*,*y*)
*v*(*x*,*y*),

*x*
*x*

*x*–*y*

= –*v*(*x*,*y*)+

–

*v*(*x*,*y*)

*x*–*y*

=

–

*v*(*x*,*y*)

*x*+*y*.

This means that, if*X*is an inner product space, inequality () is also equivalent to
inequal-ity (). Thus in an inner product space equations () and () are equivalent.

**Remark ** Inequality () can be written equivalently on the form

*v*(*x*,*y*)≤ *x*+*y*

*x*+*y*, *x*= ,*y*= . ()

By changing*y*by –*y*in () and taking into account that*α*[*x*,*y*] =*v*(*x*, –*y*), we see that
we have the Dunkl-Williams inequality in an inner product space; see [, ].

Inverse inequalities are given in the next theorem.

**Theorem ** *Let*(*X*,·,· )*be an inner product space*,*with norm* · *and let nonzero *
*ele-ments x*,*y*∈*X be such that x*+*y*= .

(i) *Ifx* ≥ *y*,*then*

*x*+*y*–*x*+*y*

≥ *x*+*y*–*v*(*x*,*y*)· *y*–*x*+*y*,*x* (*x*–*y*)

*x*+*y* · *x* ()

*and*

*x*+*y*–*x*+*y*

≥ *x*+*y*–*v*(*x*,*y*)· *x*+*x*+*y*,*y*(*x*–*y*)

*x*+*y* · *y* . ()

(ii) *Without conditionx* ≥ *y*,

*x*+*y*–*x*+*y* ≥

–

*v*(*x*,*y*)

*x*+*y*

+

*v*(*x*,*y*)

–

_{(}

*x*–*y*)

*Proof* (i) We can apply Theorem . From equation () we obtain

*x*+*y*–*x*+*y*

≥ –*v*(*x*,*y*)*y*+

– *x*,*y* +*x*

*x*+*y* · *x*

*x*–*y*

=*x*+*y*–*v*(*x*,*y*)· *y*–*x*+*y*,*x*(*x*–*y*)
*x*+*y* · *x* .

So we proved inequality ().

We apply Theorem . From equation () we deduce that

*x*+*y*–*x*+*y*

≥ –*v*(*x*,*y*)*x*–

–*x*,*y* +*y*

*x*+*y* · *y*

*x*–*y*

=*x*+*y*–*v*(*x*,*y*)· *x*+*x*+*y*,*y* (*x*–*y*)
*x*+*y* · *y* .

So we proved inequality () too.

(ii) By virtue of the symmetry of equation () we can suppose that*x* ≥ *y*. If we add
inequalities () and () and then we divide by we obtain

*x*+*y*–*x*+*y*

≥ *x*+*y*–

*v*(*x*,*y*)*x*+*y*

+ *x*,*y*(*x*–*y*)

*x* · *y* · *x*+*y*–

(*x*–*y*)

*x*+*y*

=

–

*v*(*x*,*y*)

*x*+*y*+

*v*(*x*,*y*)

–

_{(}

*x*–*y*)

*x*+*y* .

**Remark ** From the proof of Theorem it follows that in an inner product space equation
() is equivalent to equation () and equation () is equivalent to equation ().

**Remark ** Inequality () has the advantage of the symmetry, but in general it does not
improve the triangle inequality.

**Remark ** We note also that inequality () is equivalent to inequality (). Indeed,
equa-tion () can be written in equivalent form,

*x* · *x*+*y*≤*x*+*x*,*yx*–*y*+*v*(*x*,*y*)· *x* · *y* · *x*+*y*.

Then

*x*+*y* *x*,*y*
*x*+*y*+

≤*v*(*x*,*y*)· *x*+*y*,

and then

If*v*(*x*,*y*)= we obtain (). If*v*(*x*,*y*) = then equation () is obvious but also equation
() is obvious, since*v*(*x*,*y*) = implies*xy*= –*yx*and then*x*+*y*=*x*–*y*, then
inequality () is reduced to an equality.

Also inequality () is equivalent to inequality (). Indeed, equation () can be written in equivalent form,

*y*+*x*,*yx*–*y*+*y* · *x*+*y*≤*v*(*x*,*y*)· *x* · *y* · *x*+*y*,

and then

*v*(*x*,*y*)≤*v*(*x*,*y*)· *x*+*y*
*x*+*y*.

From this, we reason similarly to in the case of equation ().

In conclusion, all relations (), (), (), (), and consequently also (), are equivalent
to each other, for *x*= , *y*= ,*x*+*y*= . From Remarks and it follows that if*X* is
an inner product space inequalities (), (), (), and () are equivalent to each other, for
*x* ≥ *y*> and*xy*= –*yx*.

**Competing interests**

The authors declare that they have no competing interests.

**Authors’ contributions**

The work was carried out in collaboration between the two authors. The authors had equal contributions in writing this article. The author NM plays the role of corresponding author. All authors read and approved the ﬁnal manuscript.

**Acknowledgements**

The authors were grateful to the referee for useful comments. Received: 10 October 2016 Accepted: 22 December 2016

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