No problem! Glad I could help :)

[Saitama]

Homework Statement

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}

Homework Equations

Cauchy-Schwarz inequality maybe?

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!

 

[Ray Vickson]

Homework Statement

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}

Homework Equations

Cauchy-Schwarz inequality maybe?

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!

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.

 

[Saitama]

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.

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

 

[I like Serena]

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

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?

 

[Saitama]

Hi ILS!

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?

\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. :(

 

[I like Serena]

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. :(

Heya.

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

 

[Saitama]

Heya.

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

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 ) had success with them.

Thanks!

 

[I like Serena]

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!

Good!

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 ) had success with them.

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?

 

[Saitama]

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

[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!

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?

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!