Theorem

A positive number plus its reciprocal is at least $2$

Algebra • Expressions • Maximum and Minimum Values

Formula

x + \frac{1}{x} \geq 2 (x > 0)

For

x > 0

x + \frac{1}{x} \geq 2

, with equality exactly at

x = 1

, since

x + \frac{1}{x} - 2 = \frac{( x - 1 ) ^{2}}{x} \geq 0

; for

x < 0

the same argument gives

x + \frac{1}{x} \leq - 2

How to use it

1Whenever a quantity and its reciprocal appear together — $sin θ$ with $cosec θ$ , $tan θ$ with $cot θ$ — the sum is at least $2$ .
2Equality needs the quantity to equal $1$ ; if the stated domain excludes that value the inequality is strict.
3Read backwards it gives the greatest value $- 2$ when the quantity is negative.

What is the least value of

tan^{2} θ + cot^{2} θ

for

0 < θ < \frac{π}{2}

Step-by-step

1
Put $x = tan^{2} θ > 0$ , so $cot^{2} θ = \frac{1}{x}$
Why: The two are reciprocals of each other.
2
$x + \frac{1}{x} \geq 2$ , with equality at $x = 1$ , i.e.\ at $θ = 4 5^{\circ}$
Why: That value of $θ$ lies in the given interval, so the bound is attained.

Answer

2

Watch out

The bound needs

x > 0

; for negative

x

the sum is at most

- 2

. Where the domain keeps

x

away from

1

the inequality is strict, and an option reading

\geq 2

is then a weaker claim than the true

> 2

Also known as

x plus 1 over x is at least 2, AM-GM with a reciprocal, minimum of x + 1/x, sin plus cosec minimum