# Finding the minimum value

1. Jun 9, 2013

### Saitama

1. The problem statement, all variables and given/known data
Let x, y, z be positive real numbers such that $x+y+z=1$. Determine the minimum value of
$$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$$

2. Relevant equations
Cauchy-Schwarz inequality maybe?

3. The attempt at a solution
Applying the Cauchy Schwarz inequality,
$$\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)\leq (1^2+4^2+9^2)\left( \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)$$
I don't see how can I solve this.
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}$$
$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$ and $x+y+z=1$
$$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{(xy+yz+zx)^2-2xyz}{x^2y^2z^2}$$
I don't think the above simplification is useful here.

Any help is appreciated. Thanks!

2. Jun 9, 2013

### Ray Vickson

This problem is easily tackled using calculus. In particular, you can use either (1) the Lagrange multiplier method (the easiest); or (2) solve for z (say) in terms of x and y from the constraint equation, then substitute that expression in place of z in the function you want to minimize---giving you two-variable unconstrained minimization problem.

3. Jun 9, 2013

### Saitama

I have never done calculus questions involving three variables. Is it not possible to do this without calculus?

4. Jun 9, 2013

### I like Serena

You can do it using the AM-GM inequality $\frac{a+b}{2} = \sqrt{ab}$ with equality only if $a=b$.

\begin{aligned}\frac 1 x + \frac 4 y + \frac 9 z &= \left(\frac 1 x + \frac 4 y + \frac 9 z\right)(x+y+z) \\
&= (1+4+9) + ... \\
&\ge (1+4+9) + ... \\
&= (1+4+9) + 2 \sqrt{1 \cdot 4} + 2 \sqrt{1 \cdot 9} + 2 \sqrt{4 \cdot 9}
\end{aligned}

Can you fill in the dots?

5. Jun 9, 2013

### Saitama

Hi ILS!!!!

$$\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)(x+y+z)=14+4\left(\frac{x+z}{y}\right)+9\left(\frac{x+y}{z}\right)+\left(\frac{y+z}{x}\right)$$

I really can't figure out how you brought those surds in the end. I tried applying AM-GM to the expanded expression but that led to me nowhere. :(

6. Jun 9, 2013

### I like Serena

Heya.

Try it for instance with $\frac y x + 4\frac x y$.

7. Jun 9, 2013

### Saitama

Ah, I think I got it.
$$\frac{y}{x}+4\frac{x}{y}\geq 2\sqrt{1\cdot 4}$$
$$\frac{z}{x}+9\frac{x}{z} \geq 2\sqrt{1\cdot 9}$$
$$9\frac{y}{z}+4\frac{z}{y} \geq 2\sqrt{4\cdot 9}$$
I think I can add the inequalities as $x,y$ and $z$ are positive real numbers.
Hence,
$$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq 36$$

Thanks ILS!!!

But how can I do it with Cauchy-Schwarz? I found this problem in a trigonometry book and the author wrote that the problem can be done in one-step using Cauchy-Schwarz. As this is a trigonometry book, the author used trigonometric substitutions to solve the problem but I am curious as to how can I do it using Cauchy-Schwarz. And anyways, I always wanted to learn this inequality because a few questions were asked in my tests and I rarely (or once :tongue2:) had success with them.

Thanks!

8. Jun 9, 2013

### I like Serena

Good!

Well, your expression is on the wrong side of the Cauchy-Schwarz inequality (and you also left out a square ).

Cauchy-Schwarz is:
$$|(a \cdot b)|^2 \le (a \cdot a) (b \cdot b)$$
For the right side, you want something like
$$\left(\frac 1x + \frac 4y + \frac 9z\right)(x+y+z)$$
Which a and b would fit?

9. Jun 9, 2013

### Saitama

[strike]Why is it wrong? I think I only forgot to put a 2 in the exponent on LHS of inequality.[/strike]

EDIT: Got it. Thanks!
If I take a as $(\sqrt{x},\sqrt{y},\sqrt{z})$ and b as $(\frac{1}{\sqrt{x}},\frac{2}{\sqrt{y}},\frac{3}{\sqrt{z}})$, it gives the right answer.

Thank you again ILS!