MHB What is the Maximum Value of k for the Given Inequality?

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Value
AI Thread Summary
The discussion focuses on finding the maximum value of k for the inequality $\dfrac{25}{x^3}+\dfrac{16}{y^3}+\dfrac{9}{z^3} \geq k$ under the constraint $x^3+y^3+z^3=1$ for positive real numbers x, y, and z. Participants explore methods such as the AM-GM inequality and the Cauchy-Schwarz inequality to derive the minimum value of the left-hand side expression. The conclusion reached is that the minimum value is 144, establishing that k can be at most 144. Additionally, conditions for equality are discussed, indicating that equality holds when the ratios of the variables are specific proportional values. The conversation effectively highlights the application of inequalities in optimizing expressions under constraints.
Albert1
Messages
1,221
Reaction score
0
$x,y,z\in R^+$
given :
$x^3+y^3+z^3=1$
find the largest value of k such that the following inequality always holds:
$\dfrac {25}{x^3}+\dfrac{16}{y^3}+\dfrac {9}{z^3} \geq k$
 
Mathematics news on Phys.org
Albert said:
$x,y,z\in R^+$
given :
$x^3+y^3+z^3=1$
find the largest value of k such that the following inequality always holds:
$\dfrac {25}{x^3}+\dfrac{16}{y^3}+\dfrac {9}{z^3} \geq k$

We can freely substitute $u=x^3,v=y^3,w=z^3$ to find:

Minimize $\dfrac {25}{u}+\dfrac{16}{v}+\dfrac {9}{w}$, where $u+v+w=1$ and $u,v,w > 0$.

Applying Lagrange multipliers, we find an extremum at $u=\dfrac 5{12},\ v=\dfrac 4{12},\ w=\dfrac 3{12}$ with value 144.
With e.g. $u$ very small, the value becomes very large, so we can tell that the extremum is indeed a minimum.

So $k=144. \qquad \blacksquare$
 
As "I like Serena's notion " :we substitute $u=x^3,v=y^3,w=z^3$
to find:

Minimize $\dfrac {25}{u}+\dfrac{16}{v}+\dfrac {9}{w}$, where $u+v+w=1$ and $u,v,w > 0$.

Using "AM-GM" inequality we have:

$\dfrac {25}{u}\times (u+v+w)+\dfrac {16}{v}\times (u+v+w)+\dfrac

{9}{w}\times (u+v+w) $

$\geq (25+16+9)+2\sqrt{25\times 16}+2\sqrt{25\times 9}+2\sqrt{16\times 9}=144=k$
 
Albert said:
As "I like Serena's notion " :we substitute $u=x^3,v=y^3,w=z^3$
to find:

Minimize $\dfrac {25}{u}+\dfrac{16}{v}+\dfrac {9}{w}$, where $u+v+w=1$ and $u,v,w > 0$.

Using "AM-GM" inequality we have:

$\dfrac {25}{u}\times (u+v+w)+\dfrac {16}{v}\times (u+v+w)+\dfrac

{9}{w}\times (u+v+w) $

$\geq (25+16+9)+2\sqrt{25\times 16}+2\sqrt{25\times 9}+2\sqrt{16\times 9}=144=k$

I don't get it.
How do you apply the AM-GM inequality?

The best I can make of it, is:

$\dfrac {25}{u}\times (u+v+w)+\dfrac {16}{v}\times (u+v+w)+\dfrac {9}{w}\times (u+v+w) \\
= (25+16+9) + \dfrac {25}{u}\times (v+w)+\dfrac {16}{v}\times (u+w)+\dfrac {9}{w}\times (u+v) \\
\ge (25+16+9) + \dfrac {25}{u}\times 2\sqrt{vw}+\dfrac {16}{v}\times 2\sqrt{uw}+\dfrac {9}{w}\times 2\sqrt{uv}$

But that does not get me anywhere. :confused:
 
I like Serena said:
I don't get it.
How do you apply the AM-GM inequality?

The best I can make of it, is:

$\dfrac {25}{u}\times (u+v+w)+\dfrac {16}{v}\times (u+v+w)+\dfrac {9}{w}\times (u+v+w) \\
= (25+16+9) + \dfrac {25}{u}\times (v+w)+\dfrac {16}{v}\times (u+w)+\dfrac {9}{w}\times (u+v) \\
\ge (25+16+9) + \dfrac {25}{u}\times 2\sqrt{vw}+\dfrac {16}{v}\times 2\sqrt{uw}+\dfrac {9}{w}\times 2\sqrt{uv}$

But that does not get me anywhere. :confused:
I think the idea is to use the inequality to deduce things like $$25\frac vu + 16\frac uv \geqslant 2\sqrt{25\times16\frac{vu}{uv}}.$$ Using that approach, you might need a bit of additional argument to show that the minimum value 144 is actually attained.
 
Opalg said:
I think the idea is to use the inequality to deduce things like $$25\frac vu + 16\frac uv \geqslant 2\sqrt{25\times16\frac{vu}{uv}}.$$ Using that approach, you might need a bit of additional argument to show that the minimum value 144 is actually attained.

Yep. That makes sense.
Thanks.
 
Albert said:
$x,y,z\in R^+$
given :
$x^3+y^3+z^3=1$
find the largest value of k such that the following inequality always holds:
$\dfrac {25}{x^3}+\dfrac{16}{y^3}+\dfrac {9}{z^3} \geq k$

I just found a different solution.

According to the Cauchy-Schwarz inequality we have for all $\mathbf a$ and $\mathbf b$ in an inner product space:
$$|(\mathbf a\cdot \mathbf b)|^2 \le (\mathbf a \cdot \mathbf a)(\mathbf b \cdot \mathbf b)$$

If we pick $$\mathbf a = \left(\frac 5{x^{3/2}}, \frac 4{y^{3/2}}, \frac 3{z^{3/2}}\right)$$ and $$\mathbf b = \left(x^{3/2}, y^{3/2}, z^{3/2}\right)$$,

we get:
\begin{array}{lcl}
|(5+4+3)|^2 &\le& \left(\frac{25}{x^3} + \frac{16}{y^3} + \frac{9}{z^3}\right) (x^3+y^3+z^3) & \\
144 &\le& \frac{25}{x^3}+\frac{16}{y^3}+\frac{9}{z^3} & \qquad \blacksquare
\end{array}
 
I like Serena said:
I just found a different solution.

According to the Cauchy-Schwarz inequality we have for all $\mathbf a$ and $\mathbf b$ in an inner product space:
$$|(\mathbf a\cdot \mathbf b)|^2 \le (\mathbf a \cdot \mathbf a)(\mathbf b \cdot \mathbf b)$$

If we pick $$\mathbf a = \left(\frac 5{x^{3/2}}, \frac 4{y^{3/2}}, \frac 3{z^{3/2}}\right)$$ and $$\mathbf b = \left(x^{3/2}, y^{3/2}, z^{3/2}\right)$$,

we get:
\begin{array}{lcl}
|(5+4+3)|^2 &\le& \left(\frac{25}{x^3} + \frac{16}{y^3} + \frac{9}{z^3}\right) (x^3+y^3+z^3) & \\
144 &\le& \frac{25}{x^3}+\frac{16}{y^3}+\frac{9}{z^3} & \qquad \blacksquare
\end{array}
... with equality if (and only if) $$\Bigl(\frac 5{x^{3/2}}, \frac 4{y^{3/2}}, \frac 3{z^{3/2}}\Bigr)$$ is a scalar multiple of $\bigl(x^{3/2}, y^{3/2}, z^{3/2}\bigr)$, from which you can quickly deduce that $(x^3,y^3,z^3) = \bigl(\tfrac5{12},\tfrac4{12},\tfrac3{12}\bigr).$
 
Back
Top