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anemone
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Let $a,\,b,\,c$ and $d$ be positive real numbers such that $(a^2+b^2)^3=c^2+d^2$, prove that $\dfrac{a^3}{c}+\dfrac{b^3}{d}\ge 1$.
my solution :anemone said:Let $a,\,b,\,c$ and $d$ be positive real numbers such that $(a^2+b^2)^3=c^2+d^2---(1)$, prove that $\dfrac{a^3}{c}+\dfrac{b^3}{d}\ge 1$.
Albert said:equality occurs when $a^3d=b^3c (\,\, that \,\ is :\ a=b\,\,and\,\, c=d)$
greg1313 said:Re: Albert's solution.
I don't think $a^3d=b^3c$ implies $a=b$ and $c=d$. Consider $2^3\cdot8$ and $4^3\cdot1$.
anemone said:My solution:
Apply the Cauchy-Schwarz inequality to the sum of $a^2+b^2$, we have:
\(\displaystyle a^2+b^2=\sqrt{a}\sqrt{a^3}+\sqrt{b}\sqrt{b^3}=\frac{\sqrt{c}\sqrt{a}\sqrt{a^3}}{\sqrt{c}}+\frac{\sqrt{d}\sqrt{b}\sqrt{b^3}}{\sqrt{d}}\le \sqrt{ac+bd}\sqrt{\frac{a^3}{c}+\frac{b^3}{d}}\)
Now we rearrange it to make \(\displaystyle \frac{a^3}{c}+\frac{b^3}{d}\) on the LHS of the inequality, we see that we get:
\(\displaystyle \frac{a^3}{c}+\frac{b^3}{d}\ge\frac{(a^2+b^2)^2}{ac+bd}\)(*)
Using again the Cauchy-Schwarz inequality to the sum of $ac+bd$ we have:
\(\displaystyle ac+bd\le\sqrt{a^2+b^2}\sqrt{c^2+d^2}\)
\(\displaystyle \begin{align*}\therefore \frac{a^3}{c}+\frac{b^3}{d}&\ge\frac{(a^2+b^2)^2}{ac+bd}\\&\ge \frac{(a^2+b^2)^2}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}}\\&\ge \frac{(a^2+b^2)^{\frac{3}{2}}}{(c^2+d^2)^{\frac{1}{2}}}\\&\ge \left(\frac{(a^2+b^2)^3}{(c^2+d^2)}\right)^{\frac{1}{2}}\\&\ge 1\,\,\,\,\,\,\,\,\text{(Q.E.D.)}\end{align*}\)
Albert said:my solution :
using $AP\geq GP$
$\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^3d+b^3c}{cd}\geq \dfrac{2a^3d}{cd}=\dfrac{2a^3}{c}---(2)$
equality occurs when $a^3d=b^3c (\,\, that \,\ is :\ a=b\,\,and\,\, c=d)$
if so $(1)$ becomes:$8a^6=2c^2$ ,or $a^3=\dfrac {c}{2}$
and $(2)$ becomes:
$\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^3d+b^3c}{cd}\geq \dfrac{2a^3}{c}=1$
anemone said:Hi Albert,
Albeit it's true that equality occurs when $a=b,\,c=d$ but I don't see how we could provide the airtight proof from $a^3d=b^3c$ and $(a^2+b^2)^3=c^2+d^2$ with the conclusion that $a=b$ and $c=d$...
The variables in the inequality are typically defined as follows:
a, b, c, and d are real numbers with c and d being positive.
a and c are the numerators of the fractions, while b and d are the denominators.
The "Algebra Challenge" serves as a reminder that algebraic manipulation and properties can be used to prove the given inequality.
Yes, the inequality can also be proven using other mathematical techniques such as calculus or geometric proofs. However, algebra is typically the most straightforward approach in this case.
Yes, one strategy that can be used is to start by simplifying each fraction and then combining them to create a single expression. Then, algebraic manipulation can be used to prove that the expression is greater than or equal to 1.
One way to check your work is to substitute specific values for the variables and see if the inequality still holds. Another method is to use a graphing calculator to graph the expression and see if it is always greater than or equal to 1.