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Find the largest value of k

Albert

Well-known member
Jan 25, 2013
1,225
$x,y,z\in R^+$
given :
$x^3+y^3+z^3=1$
find the largest value of k such that the following inequality always holds:
$\dfrac {25}{x^3}+\dfrac{16}{y^3}+\dfrac {9}{z^3} \geq k$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
$x,y,z\in R^+$
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 $k=144. \qquad \blacksquare$
 

Albert

Well-known member
Jan 25, 2013
1,225
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) $

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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:
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
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 \(\displaystyle 25\frac vu + 16\frac uv \geqslant 2\sqrt{25\times16\frac{vu}{uv}}.\) Using that approach, you might need a bit of additional argument to show that the minimum value 144 is actually attained.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I think the idea is to use the inequality to deduce things like \(\displaystyle 25\frac vu + 16\frac uv \geqslant 2\sqrt{25\times16\frac{vu}{uv}}.\) 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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
$x,y,z\in R^+$
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 \(\displaystyle \mathbf a = \left(\frac 5{x^{3/2}}, \frac 4{y^{3/2}}, \frac 3{z^{3/2}}\right)\) and \(\displaystyle \mathbf b = \left(x^{3/2}, y^{3/2}, z^{3/2}\right)\),

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}
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
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 \(\displaystyle \mathbf a = \left(\frac 5{x^{3/2}}, \frac 4{y^{3/2}}, \frac 3{z^{3/2}}\right)\) and \(\displaystyle \mathbf b = \left(x^{3/2}, y^{3/2}, z^{3/2}\right)\),

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) \(\displaystyle \Bigl(\frac 5{x^{3/2}}, \frac 4{y^{3/2}}, \frac 3{z^{3/2}}\Bigr)\) is a scalar multiple of $\bigl(x^{3/2}, y^{3/2}, z^{3/2}\bigr)$, from which you can quickly deduce that $(x^3,y^3,z^3) = \bigl(\tfrac5{12},\tfrac4{12},\tfrac3{12}\bigr).$