Резюме. The following three inequalities are considered in the paper \(h_{a}+h_{b}+h_{c} \leq 4 R+r, h_{a}+h_{b}+h_{c} \leq \tfrac{a^{2}+b^{2}+c^{2}}{2 R}, h_{a}+h_{b}+h_{c} \leq \tfrac{2 s^{2}}{3 R}\). Arrangements of inequalities are considered too.

Ключови думи: triangle, altitude, inequality, generalization, arrangement

We will prove, that:

(1) \(h_{a}+h_{b}+h_{c} \leq 4 R+r \) ,

(2) \(h_{a}+h_{b}+h_{c} \leq \cfrac{a^{2}+b^{2}+c^{2}}{2 R} \) ,

(3) \( h_{a}+h_{b}+h_{c} \leq \cfrac{2 s^{2}}{3 R}\) ,

where \(a, b, c\) a are the sides of a triangle \(\triangle A B C ; h_{a}, h_{b}, h_{c}\) a are the corresponding altitudes; \(s\) is the semiperimeter and \(R\) and \(r\) are the circumradius and the inradius of the triangle. For the proof of (1) we will prove the inequality:

(4) \[ 4 R+r \geq s \sqrt{3} \]

What follows is the proof of (4).

Proof: We will use the well-known formulae (see (Grozdev, 2007)):

\[ r_{a}=\tfrac{F}{s-a}, r_{b}=\tfrac{F}{s-b}, r_{c}=\tfrac{F}{s-c} \]

where \(r_{a}, r_{b}, r_{c}\) a are the exradii of \(\triangle A B C\) and \(F\) is its area. Using that \(F=r s\) and applyingthe Heron’s formula \(F=\sqrt{s(s-a)(s-b)(s-c)}\) we get:

\[ r_{a} r_{b}=\tfrac{F^{2}}{(s-a)(s-b)}=\tfrac{s(s-a)(s-b)(s-c)}{(s-a)(s-b)}=s(s-c) \] Analogously, \(r_{b} r_{c}=s(s-a)\) and \(r_{a} r_{c}=s(s-b)\). Consequently:

\(r_{a} r_{b}+r_{b} r_{c}+r_{a} r_{c}=s(s-a)+s(s-b)+s(s-c)=s(s-a+s-b+s-c)=s(3 s-2 s)=s^{2}\) , i.e.

(5) \(r_{a} r_{b}+r_{b} r_{c}+r_{a} r_{c}=s^{2}\) .

Further, apply the well-known formula \(a b c=4 R F\) :

\(\begin{aligned} & r_{a}+r_{b}+r_{c}-r=\cfrac{F}{s-a}+\cfrac{F}{s-b}+\cfrac{F}{s-c}-\cfrac{F}{s}= \\ = & F\left[\cfrac{s-b+s-a}{(s-a)(s-b)}+\cfrac{s-s+c}{s(s-c)}\right]=F\left[\cfrac{2 s-(a+b)}{(s-a)(s-b)}+\cfrac{c}{s(s-c)}\right]= \\ = & F\left[\cfrac{c}{(s-a)(s-b)}+\cfrac{c}{s(s-c)}\right]=c F \cdot \cfrac{s(s-c)+(s-a)(s-b)}{s(s-a)(s-b)(s-c)}= \\ = & c F \cdot \cfrac{2 s^{2}-(a+b+c) s+a b}{F^{2}}=c \cdot \cfrac{a b}{F}=\cfrac{a b c}{F}=\cfrac{4 R F}{F}=4 R, \text { i.e. } \end{aligned}\)

(6) \( r_{a}+r_{b}+r_{c}=4 R+r \) .

Now, we go back to the inequality (4). Apply the following evident inequality:

\[ \begin{aligned} & \left(r_{a}+r_{b}+r_{c}\right)^{2} \geq 3\left(r_{a} r_{b}+r_{b} r_{c}+r_{a} r_{c}\right) \\ & \left(\Leftrightarrow \tfrac{1}{2}\left[\left(r_{a}-r_{b}\right)^{2}+\left(r_{b}-r_{c}\right)^{2}+\left(r_{c}-r_{a}\right)^{2}\right] \geq 0\right) \end{aligned} \]

It follows from the equalities (5) and (6) that:

\[ (4 R+r)^{2} \geq 3 \cdot s^{2} \] and from here:

\(4 R+r \geq s \sqrt{3}\), q.e.d.

An equality holds in (4) iff \(\triangle A B C\) is equilateral.

For the proof of the inequality (1) we will use also the well-known Euler's inequality:

(7) \[ R \geq 2 r \]

(Different proofs of this inequality could be found in (Arslanagić, 2005), (Arslanagić, 2008) and (Arslanagić, 2009).)

Also, we will use the following well-known equality:

(8) \[ a b+b c+c a=s^{2}+r^{2}+4 R r \]

the proof of which could be found in (Arslanagić, 2005) too.

We have from (8) that:

\(h_{a}+h_{b}+h_{c}=\cfrac{2 F}{a}+\cfrac{2 F}{b}+\cfrac{2 F}{c}=2 F\left(\cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{c}\right)= \)

\( =2 \cdot \cfrac{a b c}{4 R} \cdot \cfrac{a b+b c+c a}{a b c}=\cfrac{s^{2}+r^{2}+4 R r}{2 R}\) , i.e.

(9) \( h_{a}+h_{b}+h_{c}=\cfrac{s^{2}+r^{2}+4 R r}{2 R} \) .

Now, using the inequality (4) we get:

(10) \[ \begin{aligned} & s \sqrt{3} \leq 4 R+r \\ & \quad \Leftrightarrow 3 s^{2} \leq 16 R^{2}+8 R r+r^{2} \end{aligned} \]

and applying the inequality (7), we have:

(11) \[ \begin{aligned} & 24 R^{2}-6 R r-3 r^{2}-\left(16 R^{2}+8 R r+r^{2}\right)= \\ = & 8 R^{2}-14 R r-4 r^{2}=2\left(4 R^{2}-7 R r-2 r^{2}\right)= \\ = & 2(4 R+r)(R-2 r) \geq 0, \text { i.e. } \\ & 16 R^{2}+8 R r+r^{2} \leq 24 R^{2}-6 R r-3 r^{2} \end{aligned} \]

It follows from (10) and (11) that:

(12) \[ \begin{aligned} 3 s^{2} & \leq 24 R^{2}-6 R r-3 r^{2} \\ & \Leftrightarrow s^{2} \leq 8 R^{2}-2 R r-r^{2} \end{aligned} \]

Finally, from (9) and (12) we get:

\(h_{a}+h_{b}+h_{c}=\cfrac{s^{2}+r^{2}+4 R r}{2 R} \leq \cfrac{8 R^{2}-2 R r-r^{2}+r^{2}+4 R r}{2 R} \) , i.e.

\[ h_{a}+h_{b}+h_{c} \leq 4 R+r \] and this is the inequality (1), q.e.d.

An equality holds in (4) iff \(h_{a}=h_{b}=h_{c} \Rightarrow a=b=c\), i.e. when \(\triangle A B C\) is equilateral.

Remark 1: We will prove the following generalization of the inequality (1):

(13) \[ h_{a}^{\alpha}+h_{b}^{\alpha}+h_{c}^{\alpha} \leq 3^{1-\alpha}(4 R+r)^{\alpha} \]

where \(\alpha \in[0,1]\) is real number.

For the purpose we consider the function \(f(x)=x^{\alpha}\), where \(x \gt 0\) and \(\alpha \in[0,1]\). Since \(f^{\prime \prime}(x)=\alpha(\alpha-1) x^{\alpha-2} \leq 0\), this function is concave. Applying Jensen's inequality and the inequality (1) we get:

\[ h_{a}^{\alpha}+h_{b}^{\alpha}+h_{c}^{\alpha} \leq 3\left[\tfrac{1}{3}\left(h_{a}+h_{b}+h_{c}\right)\right]^{\alpha} \leq 3\left[\tfrac{1}{3}(4 R+r)\right]^{\alpha}=3^{l-\alpha}(4 R+r)^{\alpha} \text {, q.e.d. } \] For \(\alpha=1\) we deduce the inequality (1) as a consequence of inequality (13).

In the sequel we deal with the proof of the inequality (2). We have

\[ \begin{aligned} & h_{a}+h_{b}+h_{c}=\tfrac{2 F}{a}+\tfrac{2 F}{b}+\tfrac{2 F}{c}=2 F\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}\right)= \\ & =2 F \cdot \tfrac{a b+b c+c a}{a b c}=\tfrac{2 a b c}{4 R} \cdot \tfrac{a b+b c+c a}{a b c}=\tfrac{a b+b c+c a}{2 R} \end{aligned} \] and from the well-known inequality

\(a b+b c+c a \leq a^{2}+b^{2}+c^{2}\left(\Leftrightarrow \cfrac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \geq 0\right)\) we obtain:

\(h_{a}+h_{b}+h_{c} \leq \cfrac{a^{2}+b^{2}+c^{2}}{2 R}\) , q.e.d.

An equality holds in (2) iff \(a=b=c\), i.e. when \(\triangle A B C\) is equilateral.

It remains to prove the inequality (3). We have:

\[ h_{a}+h_{b}+h_{c}=\tfrac{2 F}{a}+\tfrac{2 F}{b}+\tfrac{2 F}{c}=2 F\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}\right)=2 F \cdot \tfrac{a b+b c+c a}{a b c} . \]

Thus, it is enough to prove that:

\[ 2 F \cdot \tfrac{a b+b c+c a}{a b c} \leq \tfrac{2 s^{2}}{3 R} \]

\[ \begin{aligned} & \Leftrightarrow 3 F R \cdot \tfrac{a b+b c+c a}{4 F R} \leq s^{2} \\ & \Leftrightarrow 3(a b+b c+c a) \leq 4 s^{2} \\ & \Leftrightarrow 3(a b+b c+c a) \leq 4 \cdot \tfrac{(a+b+c)^{2}}{4} \\ & \Leftrightarrow 3(a b+b c+c a) \leq(a+b+c)^{2} \\ & \Leftrightarrow a^{2}+b^{2}+c^{2} \geq a b+b c+c a \\ & \Leftrightarrow \tfrac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \geq 0 \end{aligned} \]

Since the last inenquality is obvious, we conclude that the inequality (3) is proved. An equality holds in (3) iff \(a=b=c\), i.e. when \(\triangle A B C\) is equilateral.

Now we will prove that the inequality (3) is stronger than the inequality (1). It is enough for the purpose to prove the inequality:

(14) \[ s^{2} \leq 4 R^{2}+4 R r+3 r^{2} \]

which is known in the mathematical literature as Gerretsen's inequality (see the papare of the present author in the same issue).

We will use the formula for the distance between the incentre and the orthocentre of \(\triangle A B C\) :

(15) \[ |I H|^{2}=2 r^{2}-4 R^{2} \cos \alpha \cos \beta \cos \gamma, \]

where \(\alpha, \beta, \gamma\) are the angles of the triangle. (Two prooofs of this formula could be found in (Arslanagić, 2005).) From the well-known the well-known identity for the angles \(\alpha, \beta, \gamma\) of \(\triangle A B C\)

\[ \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2(1+\cos \alpha \cos \beta \cos \gamma) \] we obtain:

\[ \cos \alpha \cos \beta \cos \gamma=\tfrac{1}{2}\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)-1 \]

Using the equality \(a^{2}+b^{2}+c^{2}=2\left(s^{2}+r^{2}-4 R r\right)\) it follows by the sine law that:

\(\cos \alpha \cos \beta \cos \gamma=\cfrac{1}{8 R^{2}}\left(a^{2}+b^{2}+c^{2}\right)-1\) , i.e.

\(\cos \alpha \cos \beta \cos \gamma=\cfrac{1}{4 R^{2}}\left(s^{2}-r^{2}-4 R r\right)-1\) , i.e.

(16) \(4 R^{2} \cos \alpha \cos \beta \cos \gamma=s^{2}-r^{2}-4 R r-4 R^{2} \)

From (15) and (16) we obtain:

\[ |I H|^{2}=2 r^{2}-\left(s^{2}-r^{2}-4 R r-4 R^{2}\right) \] and from here (because \(|I H|^{2} \geq 0\) ) it follows that

\[ \begin{aligned} & 2 r^{2}-\left(s^{2}-r^{2}-4 R r-4 R^{2}\right) \geq 0, \text { i.e. } \\ & s^{2} \leq 4 R^{2}+4 R r+3 r^{2} \end{aligned} \] which is the inequality (14), q.e.d.

An equality holds in (14) iff \(\alpha=\beta=\gamma=\tfrac{\pi}{3} \Rightarrow a=b=c\), i.e. when \(\triangle A B C\) is equilateral.

Further we will prove the inequality:

(17) \[ \begin{aligned} & \tfrac{a^{2}+b^{2}+c^{2}}{2 R} \leq 4 R+r \\ & \Leftrightarrow a^{2}+b^{2}+c^{2} \leq 2 R(4 R+r) \end{aligned} \]

dnd by \(a^{2}+b^{2}+c^{2}=2\left(s^{2}-r^{2}-4 R r\right)\) we have: \[ \tfrac{a^{2}+b^{2}+c^{2}}{2 R} \leq 4 R+r \]

(18) \[ \Leftrightarrow a^{2}+b^{2}+c^{2} \leq 2 R(4 R+r) \]

Since

(19) \[ \begin{aligned} & 4 R^{2}+4 R r+3 r^{2} \leq 4 R^{2}+5 R r+r^{2} \\ & \Leftrightarrow R r \geq 2 r^{2} \end{aligned} \]

\(\Leftrightarrow R \geq r\) (Euler’s inequality),

We conclude easily that the inequality (2) is stronger than the inequality (1), q.e.d.

Finally we will prove that the inequality (3) is stronger than the inequality (2), i.e. that it is valid the following inequality:

(20) \[ \begin{aligned} & \tfrac{2 s^{2}}{3 R} \leq \tfrac{a^{2}+b^{2}+c^{2}}{2 R} \\ & \quad \Leftrightarrow 4 s^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right) \\ & \quad \Leftrightarrow 4 \cdot \tfrac{(a+b+c)^{2}}{4} \leq 3\left(a^{2}+b^{2}+c^{2}\right) \\ & \quad \Leftrightarrow(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right) \\ & \quad \Leftrightarrow a^{2}+b^{2}+c^{2} \geq a b+b c+c a \\ & \quad \Leftrightarrow \tfrac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \geq 0 \end{aligned} \]

The last inequality is true and this ends the proof of the assertion that the inequality (20) is trues, i.e. that the inequality (3) is stronger than inequality (2). Consequently we have:

\[ h_{a}+h_{b}+h_{c} \leq \tfrac{2 s^{2}}{3 R} \leq \tfrac{a^{2}+b^{2}+c^{2}}{2 R} \leq 4 R+r \]

More inequalities concerning the altitudes of \(\triangle A B C\) could be found in (Bottema et al., 1969), namely:

\[ \begin{gathered} h_{a}+h_{b}+h_{c} \leq s \sqrt{3},(6.1, \text { p. } 60) \\ h_{a}+h_{b}+h_{c} \leq \tfrac{9}{2} R,(6.10, \text { p. } 62) \\ h_{a}+h_{b}+h_{c} \leq 3(R+r),(6.11, \text { p. } 62) \\ h_{a}+h_{b}+h_{c} \leq 2 R+5 r,(6.12, \text { p. } 62) \\ h_{a}+h_{b}+h_{c} \leq \tfrac{2(R+r)^{2}}{R},(6.13, \text { p. } 63) \end{gathered} \]

Including the already proved inequalities (1), (2) and (3) we obtain the following series of inequalities:

\[ h_{a}+h_{b}+h_{c} \leq \tfrac{2 s^{2}}{3 R} \leq \tfrac{2(R+r)^{2}}{R} \leq 2 R+5 r \leq \tfrac{a^{2}+b^{2}+c^{2}}{2 R} \leq s \sqrt{3} \leq 3(R+r) \leq 4 R+r \leq \tfrac{9}{2} R \text {. } \] In the series equalities hold iff and only if \(\triangle A B C\) is equilateral.

REFERENCES

Grozdev, S. (2007). For High Achievements in Mathematics. The Bulgarian Experience (Theory and Practice). Sofia: ADE. (ISBN 978-954-92139-1-1), 295 pages

Arslanagić, Š. (2005). Matematika za nadarene. Sarajevo: Bosanska riječ.

Arslanagić, Š. (2008). Matematička čitanka. Sarajevo: Grafičar promet d.o.o.

Arslanagić, Š. (2009). Matematička čitanka 1. Sarajevo: Grafičar promet d.o.o.

Bencze, M. & Š. Arslanagić (2008). A Mathematical Problem Book. Sarajevo: Grafičar promet d.o.o.

Bottema, O. et al. (1969). Geometric Inequalities. Groningen: Wolters-Noordhoff Publishing House.