SOLUTION TO PROBLEM R-4

Consider the inequality

$$\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{z} + \frac{y}{x} + \frac{z}{y} \leqno(1)$$

Multiplying both sides of (1) by 2, (1) becomes equivalent to

$$2\frac{x^2}{y^2} + 2\frac{y^2}{z^2} + 2\frac{z^2}{x^2} \ge 2\frac{x}{z} + 2\frac{y}{x} + 2\frac{z}{y} \leqno(2)$$

which is equivalent to

$$\frac{x^2}{y^2} - 2\frac{x}{z} + \frac{y^2}{z^2} + \frac{y^2}{z^2... ...c{z^2}{x^2} + \frac{z^2}{x^2} - 2\frac{z}{y} + \frac{x^2}{y^2} \ge 0 \leqno(3)$$

But (3) can be written as

$$(\frac{x}{y} - \frac{y}{z})^2 + (\frac{y}{z} - \frac{z}{x})^2 + (\frac{z}{x} - \frac{x}{y})^2 \ge 0$$

which is an evident inequality since the square of a real number is a nonnegative number.
Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Sun Feb 10 15:30:22 CST 2002