Thinking about Algebra

a \ge 2$?
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Ref: R7354

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Solution

Given that $a > 0$, $b > 0$, prove that \[\begin{equation*} \frac{a}{b} + \frac{b}{a} \ge 2. \end{equation*}\]

We can write the sum of fractions as a single fraction over the common denominator $ab$: \[\frac{a}{b} + \frac{b}{a} = \frac{a^2+b^2}{ab}.\] As $a > 0$ and $b > 0$, we can multiply our inequality by $ab>0$, so \[\begin{align*} \frac{a}{b} + \frac{b}{a} \ge 2 &\iff \frac{a^2+b^2}{ab} \ge 2 \\ &\iff a^2 + b^2 \ge 2ab \\ &\iff a^2 - 2ab + b^2 \ge 0 \\ &\iff (a - b)^2 \ge 0. \end{align*}\]

As the final inequality is always true, the first inequality must also be true.

Given that $x > 0$, $y > 0$, $z > 0$ and that $x + y + z = 3$, prove that \[\begin{equation*} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge 3. \end{equation*}\]

From the first part, if we take $a=1$ and $b=x$, we obtain \[\begin{equation*} \frac{1}{x} + x \ge 2 \end{equation*}\] and similarly with $y$ and $z$ replacing $x$. Thus, \[\begin{equation*} \frac{1}{x} + x + \frac{1}{y} + y + \frac{1}{z} + z \ge 6. \end{equation*}\]

We were told to assume that $x + y + z = 3$, so we can rewrite the above as \[\begin{equation*} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3 \ge 6, \end{equation*}\] so subtracting $3$ gives us our desired result: \[\begin{equation*} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge 3. \end{equation*}\]

There are three common means for $n$ positive numbers, $x_1 + \dotsb + x_n$. We have;

the Arithmetic Mean = $\dfrac{x_1 + \dotsb + x_n}{n}$, the Geometric Mean = $\sqrt[n]{x_1\dotsb x_n}$, and the Harmonic Mean, $\dfrac{n}{\dfrac{1}{x_1} + \dotsb + \dfrac{1}{x_n}}.$

Now it’s always true that $AM \geq GM \geq HM$. The first part of the question here is the $AM \geq GM$ inequality for the numbers $\dfrac{a}{b}$ and $\dfrac{b}{a}$.

Since the harmonic mean is less than or equal to the arithmetic mean, we have \[\begin{equation*} \frac{n}{\frac{1}{x_1} + \dotsb + \dfrac{1}{x_n}} \le \dfrac{1}{n} \left( x_1 + \dotsb + x_n \right). \end{equation*}\] We can rearrange this to become \[\begin{equation*} \frac{n^2}{x_1 + \dotsb + x_n} \le \frac{1}{x_1} + \dotsb + \frac{1}{x_n}. \end{equation*}\]

So the result we proved in the second part of the problem was a special case of this inequality.

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Thinking about Algebra

Given that $a, b > 0$, can we prove that $a/b + b/a \ge 2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

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Given that $a, b > 0$, can we prove that $a/b + b/a \ge 2$?
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