# The Power of AMGM Inequality

Considering the following function:

f(x)=x+1xx∈R+f(x) = x + \frac{1}{x} \quad\quad x \in \mathbb{R^+}

Find the minimal result of this function (without using calculus).

# Pure Evil Solution (Calculus)

I mean who will stop us from using calculus? We know that extremum of the function will be at a place where the derivative of that function equals 0.

f′(x)=(x+1x)′=(x)′+(1x)′=1−1x2f'(x) = \left(x + \frac{1}{x}\right)' = \left(x\right)' + \left(\frac{1}{x}\right)' = 1 - \frac{1}{x^2}

Substituting:

f′(x)=0  ⟹  1−1x2=0  ⟹  1x2=1  ⟹  x2=1  ⟹  x=1∨x=−1∉R+  ⟹  x=1\begin{aligned} &f'(x) = 0 \\ &\implies 1 - \frac{1}{x^2} = 0 \\ &\implies \frac{1}{x^2} = 1 \\ &\implies x^2 = 1 \\ &\implies x = 1 \lor x = -1 \notin \mathbb{R^+} \\ &\implies x = 1 \end{aligned}

from that we know that we can calculate

ymin=f(1)=1+11=2y_{min} = f(1) = 1 + \frac{1}{1} = 2

which means that fmin=(1,2)f_{min} = \left(1, 2\right)

# AMGM

From AMGM inequality we have:

12(1+1x)⩾AMGMx1xx+1x⩾2  ⟹  ymin=2\begin{aligned} &\begin{array}{rcl} \frac{1}{2}\left(1 + \frac{1}{x}\right) & \stackrel{\text{AMGM}}{\geqslant} & \sqrt{x \frac{1}{x}} \\[0.35cm] x + \frac{1}{x} & \geqslant & 2 \end{array} \\[2em] &\implies y_{min} = 2 \end{aligned}

Now we can solve for xminx_{min}

x+1x=2x2+1=2xx2−2x+1=0  ⟹  xmin=1\begin{aligned} x + \frac{1}{x} &= 2 \\ x^2 + 1 &= 2x \\ x^2 - 2x + 1 &= 0 \\[0.4em] \implies x_{min} = 1 \end{aligned}

So the minimum of function ff is

fmin=(1,2)f_{min} = \left(1, 2\right)