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Prove (a³/c) + (b³/d) ≥ 1

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anemone

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Feb 14, 2012
3,755
If $a, b, c, d >0$ and $c^2+d^2=(a^2+b^2)^3$, then show that \(\displaystyle \frac{a^3}{c}+\frac{b^3}{d} \ge 1\).
 

Jester

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MHB Math Helper
Jan 26, 2012
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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.





 
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anemone

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Feb 14, 2012
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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 \(\displaystyle x=\left(\frac{a^3}{c} \right)^{\frac{2}{3}}\) and \(\displaystyle y=\left(\frac{b^3}{d} \right)^{\frac{2}{3}}\), we see that

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

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

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

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

Hence

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

\(\displaystyle x^{\frac{3}{2}}+y^{\frac{3}{2}} \ge 1\)

\(\displaystyle \therefore \frac{a^3}{c}+\frac{b^3}{d} \ge 1\) (Q.E.D.)