Šefket Arslanagić

Professor Doctor in Mathematics
Faculty of Mathematics and Natural Sciences
University of Sarajevo
Sarajevo Bosnia and Herzegovina
E-mail: asefket@pmf.unsa.ba

Резюме. A recently published algebraic inequality is discussed and three new proofs of it are proposed. The paper is of methodological character.

Ключови думи: inequality, problem solving, proof.

In the Bulgarian journal „Matematika plus“, vol. 20 (80), Nr. 4, 2012 the next problem M+463 (by Jonuc Nedelku and Konstantin Šeu from Ploešti, Romania) is publshed:

Problem. If \(x \gt 0, y \gt 0, z \gt 0\) and \(x y z=1\), prove that:

(1) \[ \tfrac{x+y}{1+x y}+\tfrac{y+z}{1+y z}+\tfrac{z+x}{1+z x} \geq 3 \]

In the sequel three proofs of this inequality are proposed.

Proof 1: After transformation (multiplying both sides by \((1+x y)(1+y z)(1+z x) \gt 0\) ), the given inequality takes the form:

(2) \[ x^{2} z+y z^{2}+x y^{2}+x^{2} y+y^{2} z+x z^{2} \geq x+y+z+x y+y z+z x . \]

The following inequalities are valid:

(3) \[ x^{2} z+y z^{2}+x y^{2} \geq x+y+z \]

and

(4) \[ x^{2} y+y^{2} z+x z^{2} \geq x y+y z+z x \]

To prove (3) we will use the arithmetic-geometric mean inequality (\(A \geq G\) ) for three positive numbers. Because \(x y z=1\), we have:

\[ \tfrac{1}{3} x^{2} z+\tfrac{1}{3} y z^{2}+\tfrac{1}{3} y z^{2} \geq 3 \cdot \sqrt[3]{\tfrac{1}{3^{3}} \cdot(x y z)^{2} \cdot z^{3}}=z \]

\[ \begin{aligned} & \tfrac{1}{3} x^{2} z+\tfrac{1}{3} x^{2} z+\tfrac{1}{3} x y^{2} \geq 3 \cdot \sqrt[3]{\tfrac{1}{3^{3}} \cdot(x y z)^{2} \cdot x^{3}}=x \\ & \tfrac{1}{3} x y^{2}+\tfrac{1}{3} x y^{2}+\tfrac{1}{3} y z^{2} \geq 3 \cdot \sqrt[3]{\tfrac{1}{3^{3}} \cdot(x y z)^{2} \cdot y^{3}}=y \end{aligned} \]

Now add the above three inequalities and obtain (3).

The inequality (4) could be proved in a similar manner. Namely, by \(A \geq G\) and \(x y z=1\) we get:

\[ \begin{aligned} & \tfrac{1}{3} x^{2} y+\tfrac{1}{3} x^{2} y+\tfrac{1}{3} y^{2} z \geq 3 \cdot \sqrt[3]{\tfrac{1}{3^{3}} \cdot x y z \cdot x^{3} y^{3}}=x y \\ & \tfrac{1}{3} y^{2} z+\tfrac{1}{3} y^{2} z+\tfrac{1}{3} x z^{2} \geq 3 \cdot \sqrt[3]{\tfrac{1}{3^{3}} \cdot x y z \cdot y^{3} z^{3}}=z y \\ & \tfrac{1}{3} x^{2} y+\tfrac{1}{3} x z^{2}+\tfrac{1}{3} x z^{2} \geq 3 \cdot \sqrt[3]{\tfrac{1}{3^{3}} \cdot x y z \cdot x^{3} z^{3}}=x z \end{aligned} \]

Again add the last three inequalities and obtain (4).

Now (2) is a consequence of (3) and (4) by adding them. The equality in (1) holds true if and only if \(x=y=z=1\).

Proof 2: The inequality (1) is equivalent to the following one (adding 3 to both sides):

\[ \tfrac{x+y}{1+x y}+1+\tfrac{y+z}{1+y z}+1+\tfrac{z+x}{1+z x}+1 \geq 6 \] and respectively to

\[ \tfrac{(x+1)(y+1)}{1+x y}+\tfrac{(y+1)(z+1)}{1+y z}+\tfrac{(z+1)(x+1)}{1+z x} \geq 6 \]

Since \(x y z=1\), we have from here that \(x y=\tfrac{1}{z}, y z=\tfrac{1}{x}, z x=\tfrac{1}{y}\) and consequently:

(5) \[ \tfrac{z(x+1)(y+1)}{z+1}+\tfrac{x(y+1)(z+1)}{x+1}+\tfrac{y(z+1)(x+1)}{y+1} \geq 6 \]

Now by \(A \geq G\) we get:

\[ \begin{aligned} & \tfrac{z(x+1)(y+1)}{z+1}+\tfrac{x(y+1)(z+1)}{x+1}+\tfrac{y(z+1)(x+1)}{y+1} \geq \\ & \geq 3 \sqrt[3]{x y z(x+1)(y+1)(z+1)}=3 \cdot \sqrt[3]{(x+1)(y+1)(z+1)} \\ & \geq 3 \sqrt[3]{2 \sqrt{x} \cdot 2 \sqrt{y} \cdot 2 \sqrt{z}}=3 \cdot \sqrt[3]{2^{3} \sqrt{x y z}}=6 \end{aligned} \]

This ends the proof.

Proof 3: We will use the substitution \(x=\tfrac{a}{b}, y=\tfrac{b}{c}, z=\tfrac{c}{a} ;(a, b, c \gt 0)\). It implies that the given inequality is equivalent to:

(6) \[ \begin{aligned} & \tfrac{a c+b^{2}}{b(a+c)}+\tfrac{a b+c^{2}}{c(a+b)}+\tfrac{b c+a^{2}}{a(b+c)} \geq 3 \\ & \Leftrightarrow \quad \tfrac{a c+b^{2}}{b(a+c)}+1+\tfrac{a b+c^{2}}{c(a+b)}+1+\tfrac{b c+a^{2}}{a(b+c)}+1 \geq 6 \\ & \Leftrightarrow \quad \tfrac{(b+c)(a+b)}{b(a+c)}+\tfrac{(b+c)(a+c)}{c(a+b)}+\tfrac{(a+b)(a+c)}{a(b+c)} \geq 6 \end{aligned} \]

Apply again the \(A \geq G\) inequality:

\[ \begin{aligned} & \tfrac{(b+c)(a+b)}{b(a+c)}+\tfrac{(b+c)(a+c)}{c(a+b)}+\tfrac{(a+b)(a+c)}{a(b+c)} \geq \\ & \geq 3 . \sqrt[3]{\tfrac{(a+b)(b+c)(a+c)}{a b c}} \geq 3 \sqrt[3]{\tfrac{2 \sqrt{a b} \cdot 2 \sqrt{b c} \cdot 2 \sqrt{a c}}{a b c}}= \\ & =3 . \sqrt[3]{\tfrac{2^{3} \sqrt{a^{2} b^{2} c^{2}}}{a b c}}=6, \text { q.e.d. } \end{aligned} \]

Thus, (6) is proved and this implies the validity of (1). The equality holds true if and only if \(a=b=c\), i.e. \(x=y=z=1\).

REFERENCES

Arslanagić, Š. (2004). Matematika za nadarene. Sarajevo: Bosanska riječ.

Grozdev, S. (2005). Preparation for European Kangaroo. Sofia: UBM (ISBN 954-8880-20-2), 220 pages (in Bulgarian).

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