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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 [1], 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 [2]) 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 [2]) 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 [3]) 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 [4] 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

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

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

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

[4] algebra precalculus – Prove that \frac{a_1^2}{a_1+b_1}+\cdots+\frac{a_n^2}{a_n+b_n} \geq \frac{1}{2}(a_1+\cdots+a_n). – Mathematics StackExchange

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