## Least value cannot be achieved without perfectly equal sharing

The result of the least value of sum of reciprocals, used in Algebra represents an important principle of **value sharing**.

We will start with the simplest example of two positive real variables.

For two positive real variables $a$ and $b$ for example, if $a+b=1$, the least value of sum of reciprocals of $a$ and $b$ will occur when the value 1 of the sum is shared equally between the two variables, that is,

$a=b=\displaystyle\frac{1}{2}$.

The least value of the sum of reciprocals will then be,

$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}=\displaystyle\frac{1}{\frac{1}{2}}+\displaystyle\frac{1}{\frac{1}{2}}=2+2=4$.

Let's prove this result using basic algebra.

### Algebraic Proof of least value of sum of reciprocals for two variables

By the **least value of sum of reciprocals principle,** for two positive real variables $a$ and $b$,

if $a+b=1$, the sum of reciprocals of the two variables, $\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}$ will be least when value 1 of the sum is shared equally between the two variables, that is, when $a=b=\displaystyle\frac{1}{2}$.

The **least value of sum of reciprocals** will then be,

$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}=\displaystyle\frac{1}{\frac{1}{2}}+\displaystyle\frac{1}{\frac{1}{2}}=2+2=4$.

Let us assume,

$a+b=1$ and $a=b=\displaystyle\frac{1}{2}$.

We will show,

For any other combination of values of $a$ and $b$ that satisfies $a+b=1$, the sum of reciprocals of $a$ and $b$ will be larger than 4.

#### Mechanism and proof of least value of sum of reciprocals for two variables

Assume $a=\displaystyle\frac{1}{2}$ and $b=\displaystyle\frac{1}{2}$ is changed by amounts $p$ and $q$ so that,

$p+q=0$, and absolute values of both $p$ and $q$ are less than $\frac{1}{2}$.

The sum of reciprocals now will be,

$\displaystyle\frac{1}{a_1}+\displaystyle\frac{1}{b_1}$, where $a_1$ and $b_1$ are the values of $a$ and $b$ changed from $\frac{1}{2}$,

$=\displaystyle\frac{1}{\frac{1}{2}+p}+\displaystyle\frac{1}{\frac{1}{2}+q}$

$=\displaystyle\frac{1+(p+q)}{\left(\displaystyle\frac{1}{2}\right)^2+\displaystyle\frac{1}{2}(p+q)+pq}$

$=\displaystyle\frac{1}{\left(\displaystyle\frac{1}{2}\right)^2+pq}$, as $p+q=0$

$=\displaystyle\frac{1}{\left(\displaystyle\frac{1}{2}\right)^2-p^2}$, substituting $q=-p$,

The denominator is less than, $\left(\displaystyle\frac{1}{2}\right)^2$, and so, the new sum of reciprocals,

$\displaystyle\frac{1}{a_1}+\displaystyle\frac{1}{b_1} \gt 4$.

As $x$ is any change amount with the primary constraint of sum of variables remaining same, it follows that for any combination of values of $a$ and $b$ other than both $\frac{1}{2}$, the changed sum of reciprocals will be greater than 4.

In other words, the sum of reciprocals will be least when $a=b=\frac{1}{2}$, and the least value will be $4$.

#### Proof of least value of sum of reciprocals for two variables for two positive real variables with any value of sum

Instead of value 1, if sum of two positive real variables $a$ and $b$ is $m$, that is, $a+b=m$, where $m$ is any positive number, the least value of sum of reciprocals of $a$ and $b$ will occur when the value $m$ of the sum is shared equally between the two variables, that is, when,

$a=b=\displaystyle\frac{m}{2}$.

The least value will be,

$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}=\displaystyle\frac{1}{\frac{m}{2}}+\displaystyle\frac{1}{\frac{m}{2}}=\frac{2}{m}+\frac{2}{m}=\frac{4}{m}$.

As before we assume for $a+b=m$, value $m$ is equally shared between the two variables,

$a=b=\frac{m}{2}$.

Instead of $\frac{1}{2}$, equally shared values are each $\frac{m}{2}$.

The sum of reciprocals is in this case,

$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}=\frac{4}{m}$.

When we reduce $a$ from $\frac{m}{2}$ by an arbitrary positive amount $x \lt \frac{m}{2}$, the same amount is to be added to the value $\frac{m}{2}$ of second variable $b$. The new values are,

$a_1=\frac{m}{2}-x$, and

$a_2=\frac{m}{2}+x$.

The sum of reciprocals will be,

$\displaystyle\frac{1}{a_1}+\displaystyle\frac{1}{b_1}=\displaystyle\frac{1}{\frac{m}{2}-x}+\displaystyle\frac{1}{\frac{m}{2}+x}$

$=\displaystyle\frac{m}{\frac{m^2}{4}-x^2}$

$=\displaystyle\frac{1}{\frac{m}{4}-\frac{x^2}{m}}$.

As $x \lt \displaystyle\frac{m}{2}$, $\displaystyle\frac{x^2}{m} \lt \displaystyle\frac{m}{4}$.

It follows,

$\displaystyle\frac{m}{4}-\displaystyle\frac{x^2}{m} \lt \displaystyle\frac{m}{4}$,

Or, $\displaystyle\frac{4}{m}=\displaystyle\frac{1}{\frac{m}{4}} \lt \displaystyle\frac{1}{\frac{m}{4}-\frac{x^2}{m}}$.

$x$ being any arbitrary number less than $\frac{m}{2}$, the equally shared value, the least value will remain to be $\frac{4}{m}$ when the value of sum $m$ is shared equally between the two variable, where $m$ is any positive real number.

Now we will revert back to the value of sum as 1 and prove the least value principle for three variables.

### Algebraic proof of least value of sum of reciprocals for three positive real variables

By the least value of sum of reciprocals principle for three positive real variables $a$, $b$ and $c$,

When $a+b+c=1$, the least value of sum of reciprocals of the three variables will occur when the value 1 of their sum is equally shared between the three variables, that is, $a=b=c=\frac{1}{3}$. The least value will be, $3+3+3=9$.

For $a+b+c=1$, let us assume as before,

$a=b=c=\frac{1}{3}$.

The sum of reciprocals is,

$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}$

$=\displaystyle\frac{1}{\frac{1}{3}}+\displaystyle\frac{1}{\frac{1}{3}}+\displaystyle\frac{1}{\frac{1}{3}}$

$=\displaystyle\frac{3}{\frac{1}{3}}=9$

Now we will introduce changes to the equally shared values of the three variables and will show that whatever way we introduce the changes, the new sum of reciprocals will be larger than the the sum of reciprocals when the value of sum is equally shared between the three.

As the total change in the three variables must be 0 so that the unit sum relation is satisfied, at least one of the change amount must be negative. Let us assume it as $x$, where, $x \lt \frac{1}{3}$.

At this point we have,

$(\frac{1}{3}-x)+(b_1+c_1+x)=1$,

Or, $(\frac{1}{3}-x)+(\frac{2}{3}+x)=1$.

So, $b_1+c_1=\frac{2}{3}+x$.

By the **principle of least value of sum of reciprocals for two variables** (which is true we know), the sum of inverses of $b_1$ and $c_1$ will then be minimum when the value of the sum is equally shared between the two, that is, minimum value of, $\displaystyle\frac{1}{b_1}+\displaystyle\frac{1}{c_1}$ will occur when $b_1=c_1=\frac{1}{3}+\frac{x}{2}$.

The minimum value of sum of reciprocals of these two changed variables will be,

$\displaystyle\frac{1}{b_1}+\displaystyle\frac{1}{c_1}=\displaystyle\frac{2}{\frac{1}{3}+\frac{x}{2}}$.

Any other value of this sum of reciprocals will increase the sum of reciprocals of three changed values of $a$, $b$ and $c$.

Adding reciprocal of changed $a$,

$\displaystyle\frac{1}{a_1}+\displaystyle\frac{1}{b_1}+\displaystyle\frac{1}{c_1}$

$=\displaystyle\frac{1}{\frac{1}{3}-x}+\displaystyle\frac{2}{\frac{1}{3}+\frac{x}{2}}$

$=\displaystyle\frac{\frac{1}{3}+\frac{x}{2}+\frac{2}{3}-2x}{\frac{1}{9}-\frac{x}{6}-\frac{x^2}{2}}$

$=\displaystyle\frac{1-\frac{3x}{2}}{\frac{1}{9}-\frac{x}{6}-\frac{x^2}{2}}$.

$=9\left(\displaystyle\frac{\frac{1}{9}-\frac{x}{6}}{\frac{1}{9}-\frac{x}{6}-\frac{x^2}{2}}\right)$.

This multiplier factor of 9 is greater than 1 as the denominator is less than the numerator for any value of $x$.

So the the sum of reciprocals of the three variables changed in any manner from balanced equally shared value of $\frac{1}{3}$ each will be larger than the sum of reciprocals for equally shared value of the variables.

### Least value of sum of reciprocals for four or more number of variables

For **four variables,** with the same approach, taking the first changed variable as changed by a negative amount and grouping the other three variables, we will have the sum of reciprocals of variables changed from equally shared value of $\frac{1}{4}$ each is,

$\displaystyle\frac{1}{a_1}+\displaystyle\frac{1}{b_1}+\displaystyle\frac{1}{c_1}+\displaystyle\frac{1}{d_1}$

$=\displaystyle\frac{1}{\frac{1}{4}-x}+\displaystyle\frac{3}{\frac{1}{4}+\frac{x}{3}}$

$=16\left(\displaystyle\frac{\frac{1}{16}-\frac{x}{6}}{\frac{1}{16}-\frac{x}{6}-\frac{x^2}{3}}\right)$.

As the second factor is less than 1, this sum will be larger than the least value 16 of sum of reciprocals for equally shared value of $\frac{1}{4}$ each.

By the same process using the result of number of variables less than 1 from, say $n$, this result can be shown to hold good for any number of variables $n$.

The general form of this interesting principle can then be stated as,

If the sum of $n$ number of positive real variables is $m$, a positive real number, the least value of sum of reciprocals will occur when the value of sum $m$ is equally shared between the $n$ number of variables as, $\frac{m}{n}$ each.

This indeed is a very remarkable balanced result firmly depending on equal value sharing.

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