MHB Can Hölder's Inequality also be applied to other optimization problems?

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion centers on proving the inequality \( \frac{a^3}{c} + \frac{b^3}{d} \ge 1 \) under the condition \( c^2 + d^2 = (a^2 + b^2)^3 \). Participants highlight the effectiveness of using Hölder's Inequality as a method to solve this problem. A proposed solution involves defining \( x \) and \( y \) in terms of \( a, b, c, \) and \( d \) and applying Hölder's Inequality to derive the necessary relationships. The conclusion confirms that the inequality holds true, demonstrating the versatility of Hölder's Inequality in optimization problems. This discussion illustrates the application of mathematical concepts to derive significant results in optimization.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
If $a, b, c, d >0$ and $c^2+d^2=(a^2+b^2)^3$, then show that $$\frac{a^3}{c}+\frac{b^3}{d} \ge 1$$.
 
Mathematics news on Phys.org
Re: Prove (a³/c)+(b³/d) ≥1

My solution
This probably can be done with some clever manipulation of inequalities but the problem does suggest that the method of Lagrange multiplier would be effective.

If we define

$F = \dfrac{a^3}{c} + \dfrac{b^3}{d} + \lambda\left(c^2+d^2-(a^2+b^2)^3 \right)$,

so
\begin{eqnarray}
F_a &=& 3\,\frac {a^2}{c}-6a\,\lambda \left( a^2+b^{2} \right) ^{2}\\
F_b &=&3\,\frac {b^2}{d}-6b\,\lambda \left( a^2+b^{2} \right) ^{2}\\
F_c &=&-\,\frac {a^3}{c^2}+2\,\lambda c\;\;\;\;\;(*)\\
F_d &=&-\,\frac {b^3}{d^2}+2\,\lambda d\\
F_{\lambda} &=&c^2+d^2-(a^2+b^2)^3
\end{eqnarray}

Setting $F_c=0$ and solving for $\lambda$ gives $\lambda= \dfrac {a^3}{2c^3}$. From the $F_a=0$ equation we have

$-3\,{\dfrac {{a}^{2} \left( -c+{a}^{3}+a{b}^{2} \right) \left( c+{a}^{
3}+a{b}^{2} \right) }{{c}^{3}}}
=0$

from which we obtain $c = a^3+ab^2$. Factoring the $F_b=0$ equation gives

$3\,{\dfrac { \left( {a}^{2}b+{b}^{3}-d \right) b}{ \left( {a}^{2}+{b}^{
2} \right) d}}=0$

leading to $d = a^2b+b^3$ noting that the remaining equation in (*) are satisfied. Thus, we have

$\dfrac{a^3}{c} + \dfrac{b^3}{d} = \dfrac{a^3}{a^3+ab^2} + \dfrac{b^3}{a^2b+b^3} = 1$

the minimum as required.
 
Re: Prove (a³/c)+(b³/d) ≥1

Thanks for participating, Jester and well done! And now it seems to me Lagrange Multiplier is a magic tool and can be applied to all kind of optimization problems to find the desired extrema points.

I want to share with you and others the solution proposed by other as well. It suggests the use of Hölder's Inequality to solve this problem.

If we let $$x=\left(\frac{a^3}{c} \right)^{\frac{2}{3}}$$ and $$y=\left(\frac{b^3}{d} \right)^{\frac{2}{3}}$$, we see that

$$a^2+b^2=c^{\frac{2}{3}}\left(\frac{a^3}{c} \right)^{\frac{2}{3}}+d^{\frac{2}{3}}\left(\frac{b^3}{d} \right)^{\frac{2}{3}}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;=c^{\frac{2}{3}}(x)+d^{ \frac{2}{3}}(y)$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\le (c^2+d^2)^{\frac{1}{3}}(x^{\frac{3}{2}}+y^{\frac{3}{2}})^{\frac{2}{3}}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\le (a^2+b^2)(x^{\frac{3}{2}}+y^{\frac{3}{2}})^{\frac{2}{3}}$$

Hence

$$a^2+b^2\le (a^2+b^2)(x^{\frac{3}{2}}+y^{\frac{3}{2}})^{\frac{2}{3}}$$

$$x^{\frac{3}{2}}+y^{\frac{3}{2}} \ge 1$$

$$\therefore \frac{a^3}{c}+\frac{b^3}{d} \ge 1$$ (Q.E.D.)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top