No Ordinary Optimization Problem

The restriction and minimized expression are of special forms $\sum_{k=1}^3{g(x_k)}=0,\\\sum_{k=1}^3{f(x_k)}=\mathrm{min.}$

with $g(x) \neq 0$ . In this case, the theorem of Lagrange multipliers takes a simple form $\lambda=\frac{f'(x_k)}{g'(x_k)}=\mathrm{const.}$

After calculating $\lambda(x)$ and its first derivative, we can see that $\lambda(x)=10\left(x+\frac{1}{x}\right)^9\left(1-\frac{1}{x^2}\right),\\ \lambda'(x)=90\left(x+\frac{1}{x}\right)^8\left(1-\frac{1}{x^2}\right)^2+20\left(x+\frac{1}{x}\right)^9\frac{1}{x^3},$ where $\lambda'(x) > 0$ for $x > 0$ , so the only way for $\lambda(x_k)$ to be constant for all $k$ is for all $x_k$ to be equal.

It follows that $x_k=a,b,c=1/3$ and the minimum of our expression is $10^{10}/3^9$ .

AM of nth powers>=AM power n: Let R=(a+1/a)^10+(b+1/b)^10+(c+1/c)^10 R/3>=[{(a+1/a)+(b+1/b)+(c+1/c)}/3]^10 Also, using AM>=HM, we get: 1/a+1/b+1/c>=9 R/3>=[{a+b+c+1/a+1/b+1/c}/3]^10>=[{1+9}/3]^10=(10/3)^10 So, R>=(10^10)/(3^9)

I don't know how to justify this solution, but I set $a+\frac{1}{a}=b+\frac{1}{b}=c+\frac{1}{c}$ , which implies $a=b=c$ . Perhaps some sort of curvature argument involving the AM-GM inequality will work?

Using lagrange multipliers, we see that a=b=c. So a b and c must be 1/3.

Take a=b=c=1/3