Cauchy and Fractional Inequalities

What this move does

The fraction form of the Cauchy inequality, $\sum \frac{x _{i}^{2}}{y _{i}} \geq \frac{( \sum x _{i} ) ^{2}}{\sum y _{i}}$ ( $y_{i} > 0$ ), is made for “a row of fractions added together”. When a numerator is not a square, first make it one:

Multiply the top and bottom of each term by a positive expression so that every term has the form $\frac{( \dots ) ^{2}}{( \dots )}$ . Choose the multiplier so that the merged numerator and denominator are easy to handle (ideally symmetric expressions or known quantities).
Merge once: $\sum \frac{x _{i}^{2}}{y _{i}} \geq \frac{( \sum x _{i} ) ^{2}}{\sum y _{i}}$ .
What remains is a polynomial inequality. Check that it holds, and that its equality case agrees with the equality condition of step 2.

Step 1 is the crux. The same fraction can be “squared up” in many ways, and each gives a different lower bound after merging: some stay above the target everywhere, others drop below it.

Predict first

We want to prove $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2}$ ( $a, b, c > 0$ ). If we write each term as $\frac{( a ) ^{2}}{b + c}$ and merge, does the proof go through?

YesNoOnly when

a, b, c

are all equal

Choose an answer first; then you can see the result.

Work through one

This is the famous Nesbitt inequality (1903).

Example: a, b, c > 0; prove a/(b+c) + b/(c+a) + c/(a+b) ≥ 3/2

Step 1 / 5

The laziest rewrite fails
The numerator is $a = (a)^{2}$ ; merge directly. At $a, b, c = 1, 2, 3$ the lower bound is about $1.43$ , already below $1.5$ .

$S \geq \frac{( a + b + c ) ^{2}}{2 ( a + b + c )} \leq \frac{3}{2}$
Multiply top and bottom by a
The polynomial inequality that is left
The two equality conditions agree
Higher powers are not always better

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: the merged lower bound is at least $\frac{3}{2}$ for all positive $a, b, c$

This route works everywhere: the reading “Lowest bound over all a, b, c” is exactly $1.5$ , reached at $a = b = c$ .

Practice

Exercise 1Basic

Positive numbers $a, b, c$ satisfy $a + b + c = 3$ . Find the minimum of $\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}$ .

Exercise 2Basic

Positive numbers $x, y, z$ satisfy $x + y + z = 1$ . Find the minimum of $\frac{x ^{2}}{y + z} + \frac{y ^{2}}{z + x} + \frac{z ^{2}}{x + y}$ .

Exercise 3Advanced

Let $a, b, c > 0$ . Prove that $\frac{a}{b + 2 c} + \frac{b}{c + 2 a} + \frac{c}{a + 2 b} \geq 1$ .

Exercise 4Challenge

(IMO 1995) Positive numbers $a, b, c$ satisfy $ab c = 1$ . Prove that $\frac{1}{a ^{3} ( b + c )} + \frac{1}{b ^{3} ( c + a )} + \frac{1}{c ^{3} ( a + b )} \geq \frac{3}{2}$ .

What this move does

Work through one

Example: a, b, c > 0; prove a/(b+c) + b/(c+a) + c/(a+b) ≥ 3/2

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

The merged lower bound is smaller than the target