Minimum of $\sum_{cyc}\frac1x$, $\sum_{cyc}x=3a>0$, $x,y,z>0$

Question

Let $f(x,y,z)=\frac1x+\frac1y+\frac1z.$ Impose the constraints $x+y+z=3a$ $x,y,z>0$. What is the minumum value of $f$?

Method of Lagrange's multipliers give $-\frac1{x^2}=-\frac1{y^2}=-\frac1{z^2}=\lambda$ and hence $x=y=z=a$ with the minumum value $\frac{3}{a}$.

How can we solve this question in a mathematics contest which does not allow derivatives?

I am stuck without calculus. Thanks in advance.

Edit: I just noticed that AM-HM inequality solves the question immediately. Can you propose another way?

Answer 1

By Cauchy–Schwarz inequality:

$$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z \right) \ge \left(\frac{1}{\sqrt{x}}\cdot \sqrt{x}+\frac{1}{\sqrt{y}}\cdot \sqrt{y}+\frac{1}{\sqrt{z}}\cdot \sqrt{z}\right)^2 = 9$$

By Jensen's inequality: The function $f(x) =\frac{1}{x}$ is convex, then

$$\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{3}\left(f(x)+f(y)+f(z) \right) \ge f\left( \frac{x+y+z}{3}\right) = \frac{1}{a}$$

Answer 2

The harmonic mean is smaller than the geometric mean which in turn is smaller than the arithmetic mean. In particular $${n\over {1\over x_1}+\ldots +{1\over x_n}}\le {x_1+\ldots +x_n\over n}$$ and the equality occurs only when $x_1=\ldots =x_n.$ For $n=3$ and $x_1+x_2+x_3=3a$ the conclusion follows.