# Chaitanya's Random Pages

## March 26, 2016

### Applying AM-GM in the denominator after flipping the sign

Filed under: mathematics — ckrao @ 8:44 pm

There are times when solving inequalities that one has a sum of fractions in which applying the AM-GM inequality to each denominator results in the wrong sign for the resulting expression.

For example (from , p18), if we wish to show that for real numbers $x_1, x_2, \ldots, x_n$ with sum $n$ that $\displaystyle \sum_{i = 1}^n \frac{1}{x_i^2 + 1}\geq \frac{n}{2},$

we may write $x_i^2 + 1 \geq 2x_i$ (equivalent to $(x_i-1)^2 \geq 0$), but this implies $\frac{1}{x_i^2 + 1} \leq \frac{1}{2x_i}$ and so the sign goes the wrong way.

A way around this is to write \begin{aligned} \frac{1}{x_i^2 + 1} &= 1 - \frac{x_i^2}{x_i^2 + 1}\\ &\geq 1 - \frac{x_i^2}{2x_i}\\ &= 1 - \frac{x_i}{2}. \end{aligned}

Summing this over $i$ then gives $\sum_{i=1}^n \frac{1}{x_i^2 + 1} \geq n - \sum_{i=1}^n (x_i/2) = n/2$ as desired.

Here are a few more examples demonstrating this technique.

2. (p9 of ) If $a,b,c$ are positive real numbers with $a + b + c = 3$, then $\dfrac{a}{1 +b^2} + \dfrac{b}{1 +c^2} + \dfrac{c}{1 +a^2} \geq \dfrac{3}{2}.$

To prove this we write \begin{aligned} \frac{a}{1 + b^2} &= a\left(1 - \frac{b^2}{1 + b^2}\right)\\ &\geq a\left(1 - \frac{b}{2}\right) \quad \text{(using the same argument as before)}\\ &=a - \frac{ab}{2}. \end{aligned}

Next we have $3(ab + bc + ca) \leq (a + b + c)^2 = 9$ as this is equivalent to $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$. This means $ab + bc + ca \leq 3$. Putting everything together, \begin{aligned} \frac{a}{1 + b^2} + \frac{b}{1 + c^2} + \frac{c}{1 + a^2}&\geq \left( a - \frac{ab}{2} \right) + \left( b - \frac{bc}{2} \right) + \left( c - \frac{ca}{2} \right)\\ &= (a + b + c) - (ab + bc + ca)/2\\ &\geq 3 - 3/2\\ &=\frac{3}{2}, \end{aligned}

as required.

3. (based on p8 of ) If $x_i > 0$ for $i= 1, 2, \ldots, n$ and $\sum_{i = 1}^n x_i^2 = n$ then $\displaystyle \sum_{i=1}^n \frac{1}{x_i^3 + 2} \geq \frac{n}{3}.$

By the AM-GM inequality, $x_i^3 + 2 = x_i^3 + 1 + 1 \geq 3x_i$, so \begin{aligned} \frac{1}{x_i^3 + 2} &= \frac{1}{2}\left( 1 - \frac{x_i^3}{x_i^3 + 2} \right)\\ &\geq \frac{1}{2}\left( 1 - \frac{x_i^3}{3x_i} \right)\\ &= \frac{1}{2}\left( 1 - \frac{x_i^2}{3} \right). \end{aligned}

Summing this over $i$ gives \begin{aligned} \sum_{i=1}^n \frac{1}{x_i^3 + 2} &\geq \frac{1}{2} \sum_{i=1}^n \left( 1 - \frac{x_i^2}{3} \right)\\ &= \frac{1}{2}\left( n - \frac{n}{3} \right)\\ &= \frac{n}{3}. \end{aligned}

4. (from ) If $x, y, z$ are positive, then $\dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } \geq \dfrac {x + y + z}{2}.$

Once again, focusing on the denominator, \begin{aligned} \dfrac {x ^ 3}{x ^ 2 + y ^ 2} &= x\left(1 - \dfrac {y ^ 2} {x ^ 2 + y ^ 2} \right)\\ &\geq x \left(1 -\dfrac{xy^2}{2xy} \right)\\ &= x-\dfrac{y}{2}. \end{aligned}

Hence, \begin{aligned} \dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } &\geq x-\dfrac{y}{2} + y-\dfrac{z}{2} + z-\dfrac{x}{2}\\ &= \dfrac {x + y + z}{2}, \end{aligned}

as desired.

5. (from the 1991 Asian Pacific Maths Olympiad, see  for other solutions) Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be positive numbers with $\sum_{i = 1}^n a_i = \sum_{i = 1}^n b_i$. Then $\displaystyle\sum_{i=1}^n\frac{a_i^2}{a_i + b_i} \geq \frac{1}{2}\sum_{i=1}^n a_i.$

Here we write \begin{aligned} \sum_{i=1}^n\frac{a_i^2}{a_i + b_i} &= \sum_{i=1}^n a_i \left(1 - \frac{b_i}{a_i + b_i} \right)\\ &\geq \sum_{i=1}^n a_i \left(1 - \frac{b_i}{2\sqrt{a_i b_i}} \right) \\ &= \frac{1}{2} \sum_{i=1}^n \left( 2a_i - \sqrt{a_i b_i} \right) \\ &= \frac{1}{4} \sum_{i=1}^n \left( 4a_i - 2\sqrt{a_i b_i} \right)\\ &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i +a_i - 2\sqrt{a_i b_i} + b_i \right) \quad \text{(as } \sum_{i=1}^n a_i = \sum_{i=1}^n b_i\text{)}\\ &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i + \left(\sqrt{a_i} - \sqrt{b_i}\right)^2 \right)\\ &\geq \frac{1}{4} \sum_{i=1}^n 2a_i\\ &= \frac{1}{2} \sum_{i=1}^n a_i, \end{aligned}

as required.

#### References

 Zdravko Cvetkovski, Inequalities: Theorems, Techniques and Selected Problems, Springer, 2012.

 Wang and Kadaveru, Advanced Topics in Inequalities, available from http://www.artofproblemsolving.com/community/q1h1060665p4590952

 Cauchy Reverse Technique: https://translate.google.com.au/translate?hl=en&sl=ja&u=http://mathtrain.jp/crt&prev=search

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