Solve this with CBS: minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$

Question

How can you see the minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ using the CBS inequality?

I have seen a proof of that that use trigonometric substitutions, but I don't see as one-step the solution using the CBS inequality.

Answer 1

Consider positive values $x,y,z$.

Denote $$(a,b,c)=\left( \frac{1}{\sqrt{x}}, \frac{2}{\sqrt{y}}, \frac{3}{\sqrt{z}}\right),$$ $$(A,B,C)=\left( \sqrt{x}, \sqrt{y}, \sqrt{z}\right),$$

then (by CBS inequality) $$(a^2+b^2+c^2)(A^2+B^2+C^2)\ge (aA+bB+cC)^2,$$ so $$ \left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \left(x+y+z\right)\ge (1+2+3)^2=36; $$ $$ \frac{1}{x}+\frac{4}{y}+\frac{9}{z}\ge 36. $$ (I have added the square)

Answer 2

You can also take a look at lagrange multiplier because it is helpful finding where the inequalties are sharp and what are the maximal minimal values when dealing with reel numbers (without the restriction of positive-negative);

So let me introduce $k$ that is the lagrange multiplier;

$f(x,y,z,k)=\dfrac1x+\dfrac4y+\dfrac9z+k(x+y+z-1)$ we will try to minimize this expression using what we know about $x,y,z$ ; Take the derivatives in respect to all variables;

$$f_k'=0$$ $$f_x'=k-\frac{1}{x^2}$$ $$f_y'=k-\dfrac{4}{y^2}$$ $$f_z'=k-\dfrac{9}{z^2}$$

All these derivatives are equal to $f_k'=0$ so we can say;

$$k=\frac{1}{x^2}=\frac{4}{y^2}=\frac{9}{z^2}$$ $$\sqrt{k}=\frac1x=\frac2y=\frac3z=\frac{1+2+3}{x+y+z}$$ $$\sqrt{k}=\frac61=6$$ So $x=\dfrac16$--$y=\dfrac26$--$z=\dfrac36=\dfrac12$

$\frac1x=6+\frac4y=12+\frac9z=18=36$ Done!