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

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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.
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If $a, b, c, d >0$ and $c^2+d^2=(a^2+b^2)^3$, then show that $$\frac{a^3}{c}+\frac{b^3}{d} \ge 1$$.
 
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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.)
 
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