Резюме. In this paper we consider a very useful inequality that Murray Klamkin \({ }^{1)}\) proved in 1975 (Uldmkin, 1975). The inequality has many applications, proving new inequalities included. A proof and some applications are proposed.

Ключови думи: Klamkin‘s inequality; triangle sides; scalar product; law of cosines; application; examples

Theorem 1. (Klamkin‘s inequality). Let \(x, y\) and \(z\) be real numbers such that \(x+y+z \gt 0\). Then for any point \(P\) in the plane of the triangle \(\triangle A B C\) the following

inequality holds true:

(1)\[ x|P A|^{2}+y|P B|^{2}+z|P C|^{2} \geq \tfrac{y z a^{2}+z x b^{2}+x y c^{2}}{x+y+z} \]

where \(a, b, c\) a are the lengts of the sides of the triangle \(\triangle A B C\).

The equality in (1) holds if and only if the point \(P\) satisfies the equality

(2)\[ \overrightarrow{A P}=\tfrac{y}{x+y+z} \overrightarrow{A B}+\tfrac{z}{x+y+z} \overrightarrow{A C} \]

Proof: Observe the vector \(x \overrightarrow{P A}+y \overrightarrow{P B}+z \overrightarrow{P C}\). Evidently, the next inequality \[ (x \overrightarrow{P A}+y \overrightarrow{P B}+z \overrightarrow{P C})^{2} \geq 0 \] holds true.

Because of the properties of the scalar product, this inequality (3) has the form

(4)\[ x^{2}|P A|^{2}+y^{2}|P B|^{2}+z^{2}|P C|^{2}+2 x y \overrightarrow{P A} \cdot \overrightarrow{P B}+2 y z \overrightarrow{P B} \cdot \overrightarrow{P C}+2 z x \overrightarrow{P C} \cdot \overrightarrow{P A} \geq 0 \]

By using the law of cosines, we get the equalities

(5)\[ \left.\begin{array}{l} 2 \overrightarrow{P A} \cdot \overrightarrow{P B}=|P A|^{2}+|P B|^{2}-c^{2} \\ 2 \overrightarrow{P B} \cdot \overrightarrow{P C}=|P B|^{2}+|P C|^{2}-a^{2} \\ 2 \overrightarrow{P C} \cdot \overrightarrow{P A}=|P C|^{2}+|P A|^{2}-b^{2} \end{array}\right\} \]

Now from (4) and (5), we obtain:

(6)\[ \left(x^{2}+x y+x z\right)|P A|^{2}+\left(y^{2}+y x+y z\right)|P B|^{2}+\left(z^{2}+z x+z y\right)|P C|^{2}-y z a^{2}-z x b^{2}-x y c^{2} \geq 0 \]

Finally, if we divide the inequality (6) by \(x+y+z \gt 0\), we get the inequality (1).

Evidently, the equality holds if and only if \(x \overrightarrow{P A}+y \overrightarrow{P B}+z \overrightarrow{P C}=\overrightarrow{0}\), i.e. \(x \overrightarrow{P A}+y(\overrightarrow{P A}+\overrightarrow{A B})+z(\overrightarrow{P A}+\overrightarrow{A C})=\overrightarrow{0}\), and from here we get (2) after arrangment.

In the sequel we propose several examples of application of the inequality (1).

Example 1. For any point \(P\) in the plane of the triangle \(\triangle A B C\) the following inequality holds true:

(7)\[ |P A|^{2}+|P B|^{2}+|P C|^{2} \geq \tfrac{a^{2}+b^{2}+c^{2}}{3} \]

Solution: This inequality follows directly from (1) when \(x=y=z=1\).

Accounting for (2), the equality holds in (7) if and only if

\[ \overrightarrow{A P}=\tfrac{1}{3} \overrightarrow{A B}+\tfrac{1}{3} \overrightarrow{A C}=\tfrac{2}{3} \overrightarrow{A A_{1}} \] where \(A_{1}\) is the middpoint of the side \(B C\). It means that the point divides the median \(A A_{l}\) in ratio \(2: 1\) computed from the vertex of the triangle, i.e. the point \(M\) is the centroid of the triangle. Example 2. In every triangle \(\triangle A B C\) the following inequality holds true:

(8)\[ a^{2}+b^{2}+c^{2} \leq 9 R^{2} \]

Solution: Put \(P \equiv O\) in inequality (1), where the point \(O\) is the circumcenter of the triangle \(\triangle A B C\). Since \(|O A|=|O B|=|O C|=R\), now it follows from (7) that

\[ \begin{aligned} & 3 R^{2} \geq \tfrac{a^{2}+b^{2}+c^{2}}{3}, \text { i.e. } \\ & a^{2}+b^{2}+c^{2} \leq 9 R^{2}, \text { q.e.d. } \end{aligned} \]

The equality holds in (8) if and only if \(a=b=c\), i.e. for equilateral triangle.

Example 3. For anyone point \(P\) in the plane of the triangle \(\triangle A B C\) the followng inequality holds true:

(9)\[ a|P A|^{2}+b|P B|^{2}+c|P C|^{2} \geq a b c . \]

Solution: The proof follows directly from (1) when \(x=a, y=b, z=c\).

Because of (2), the equality in (9) holds if and only if

\[ \overrightarrow{A P}=\tfrac{b}{a+b+c} \overrightarrow{A B}+\tfrac{c}{a+b+c}=\tfrac{b c}{a+b+c}\left(\tfrac{\overrightarrow{A B}}{c}+\tfrac{\overrightarrow{A C}}{b}\right) \]

It follows now that the vector \(\overrightarrow{A P}\) is collinear with the angular bisector. Analogously, it follows that the vectors \(\overrightarrow{B P}\) and \(\overrightarrow{C P}\) are collinear with the corresponding angular bisectors. Therefore, \(P \equiv I\), where \(I\) is the incenter.

Example 4. (Euler’s inequality) In every triangle \(\triangle A B C\) the following ine quality holds true:

(10)\[ R \geq 2 r . \]

Solution: Let \(P \equiv O\), where \(O\) is the circumcenter, i.e. \(|P A|=|O A|=R\), \(|P B|=|O B|=R\) and \(|P C|=|O C|=R\). Now it follows from (9) that

\[ \begin{aligned} & R^{2}(a+b+c) \geq a b c \\ & \Rightarrow R^{2} \geq \tfrac{a b c}{a+b+c} \end{aligned} \] and from here using the formulas \(a b c=4 R F=4 R r s\) and \(a+b+c=2 s\) we obtain:

\[ \begin{gathered} R^{2} \geq \tfrac{4 R r s}{2 s}, \text { i.e. } \\ R \geq 2 r, \text { q.e.d. } \end{gathered} \]

The equality in (10) holds for \(a=b=c\), i.e. for the equilateral triangle.

Example 5. For any point \(P\) in the plane of the triangle \(\triangle A B C\) the followng inequality holds true:

(11)\[ \sin 2 \alpha|P A|^{2}+\sin 2 \beta|P B|^{2}+\sin 2 \gamma|P C|^{2} \geq 2 F \]

Solution: Put \(x=\sin 2 \alpha, y=\sin 2 \beta, z=\sin 2 \gamma\) in (1). The right hand side of the inequality (1) takes the form:

\[ \tfrac{a^{2} \sin 2 \beta \sin 2 \gamma+b^{2} \sin 2 \gamma \sin 2 \alpha+c^{2} \sin 2 \alpha \sin 2 \beta}{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma} \]

Applying the law of sines formulas \(\sin 2 \alpha=2 \sin \alpha \cos \alpha, \sin 2 \beta=2 \sin \beta \cos \beta\), \(\sin 2 \gamma=2 \sin \gamma \cos \gamma\)

\(\cfrac{16 R^{2} \sin \alpha \sin \beta \sin \gamma(\sin \alpha \cos \beta \cos \gamma+\cos \alpha \sin \beta \cos \gamma+\cos \alpha \cos \beta \sin \gamma)}{\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma} \\ \text{and the identities}\)

\(\sin 2\alpha +\sin 2\beta + \sin 2\gamma=4\sin\alpha\sin\beta\sin\gamma\)

and

\[ \sin \alpha \cos \beta \cos \gamma+\cos \alpha \sin \beta \cos \gamma+\cos \alpha \cos \beta \sin \gamma=\sin \alpha \sin \beta \sin \gamma \] the right hand side of the inequality (1) takes the form

\[ 4 R^{2} \sin \alpha \sin \beta \sin \gamma \]

Finally, observe that

\[ 4 R^{2} \cdot \tfrac{a}{2 R} \cdot \tfrac{b}{2 R} \cdot \tfrac{c}{2 R}=\tfrac{a b c}{2 R}=2 F \]

Consequently, the inequality (11) is true because it follows from the inequality (1).

The equality in (11) holds if and only if \(P \equiv O\), which is left from for the reader to prove it.

Example 6. Let \(P\) be an arbitrary point in the interior of the trinagle \(\triangle A B C\). Prove the inequality

(12)\[ \tfrac{|P A|^{2}}{c}\left(\tfrac{1}{a}+\tfrac{1}{b}\right)+\tfrac{|P B|^{2}}{a}\left(\tfrac{1}{b}+\tfrac{1}{c}\right)+\tfrac{|P C|^{2}}{b}\left(\tfrac{1}{c}+\tfrac{1}{a}\right) \geq 2 \]

Solution: This inequality is evidently equivalent to the inequality

(13)\[ (a+b)|P A|^{2}+(b+c)|P B|^{2}+(a+c)|P C|^{2} \geq 2 a b c \]

We will now use the inequality (1).

If \(x=a, y=b, z=c\) and \(x=b, y=c, z=a\), , we obtain two inequalities:

\[ a|P A|^{2}+b|P B|^{2}+c|P C|^{2} \geq a b c \]

and

\[ b|P A|^{2}+c|P B|^{2}+a|P C|^{2} \geq a b c \] Summing the two inequalites, we obtain the following inequality

\[ |P A|^{2}(a+b)+|P B|^{2}(b+c)+|P C|^{2}(a+c) \geq 2 a b c, \] Which in fact is the inequality (13), thas proving (12).

The equality holds in (12) if and only if is \(a=b=c\), i.e. for equilateral triangle.

NOTES

1. Murray Klamkin (1921 – 2004) is a Canadian mathematician, born in USA

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