# Chaitanya's Random Pages

## March 28, 2016

### Recent months of global warmth

Filed under: climate and weather — ckrao @ 5:02 am

According to both NASA’s Goddard Institute for Space Studies [1] and the NOAA National Centers for Environmental Information [2], February 2016 set another record of the highest deviation of global temperatures above the monthly mean. In fact NASA’s dataset has seen the past five months record the largest five monthly global warm anomalies [3]. Some plots of global temperatures from recent months can be seen at Makiko Sato’s page here. One case in point is Longyearbyen, Svalbard (78°N) whose temperatures have barely been below average for the past six months (data from [4-5]).

February set the record of greatest anomaly from mean monthly temperatures, beating the previous record (set only the previous month) by more than 0.2°C. The map here shows that the vast majority of the planet had above-average temperatures, with the greatest deviation in the arctic region. As an example, check out the temperatures of Salekhard, Russia on the arctic circle during this time (this is a place that registers temperatures below -40 during winters). Over the month its average was 12.5°C above the mean! Data is from [6].

[2] NOAA National Centers for Environmental Information, State of the Climate: Global Analysis for February 2016, published online March 2016, retrieved on March 27, 2016 from http://www.ncdc.noaa.gov/sotc/global/201602.

## 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.