Prove that if $\frac ab=\frac bc=\frac cd$, then $\frac ad=\frac {a^3+b^3+c^3}{b^3+c^3+d^3}$

Question

If $a,b,c,d$ are in continued proportion, prove that $a:d=a^3+b^3+c^3:b^3+c^3+d^3$

This means $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$ and $\dfrac{a}{d}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}$

$\Rightarrow \dfrac{a}{d}-1=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}-1$

$\Rightarrow \dfrac{a-d}{d}=\dfrac{a^3-d^3}{b^3+c^3+d^3}$

$\Rightarrow \dfrac{a-d}{d}=\dfrac{(a-d)(a^2+ad+d^2)}{b^3+c^3+d^3}$

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

$\Rightarrow a^2d+ad^2+d^3=b^3+c^3+d^3$

$\Rightarrow a^2d+ad^2=b^3+c^3=(b+c)(b^2-bc+c^2)$

From $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$, $ad=bc$ and $b^2=ac$ and $c^2=bd$, hence $\Rightarrow abc+bcd=(b+c)(ac-bc+bd)$

$\Rightarrow abc+bcd=abc-b^2c+b^2d+ac^2-bc^2+bcd$

$\Rightarrow 0=ac^2-bc^2-b^2c+b^2d$

$\Rightarrow 0=b^2c-bc^2-b^2c+bc^2$

$\Rightarrow 0=0$

So is this a proof? I essentially just did a calculation and plugged in values and got everything equals zero in the end. Not sure how this proves anything.

Answer 1

You are correct about having doubt. You have taken the proposition $P$ which you want to prove, and written $P\implies ... \implies ... \implies 0=0.$ What you should have is $0=0\implies ... \implies ... \implies P,$ but it is impossible to know how to begin. So instead, start with $P$ and deduce things that are $equivalent$ to $P.$ So if you can get $P\iff ... \iff ... \iff 0=0,$ then you have $0=0\implies P.$

In your proof, you can simply replace the first 3 instances of $\implies$ with $\iff.$ Then you have the line $ \dfrac{a-d}{d}=\dfrac{(a-d)(a^2+ad+d^2)}{b^3+c^3+d^3},$ which is not equivalent to the next line unless $a\ne d.$ However it is equivalent to $0=0$ if $a=d.$

I find your other algebraic work correct, and ingenious, but there is a simpler way: Let $a/b=b/c=c/d=x.$ So $a=dx^3, b=dx^2, c=dx.$ Substitute these values for $a,b,c, $ in terms of $d$ and $x$, into proposition $P,$ and see what happens.

BTW you can use \implies for $\implies$, which makes it easier to read your own code. The LaTex for $\iff$ is \iff.

Answer 2

In this answer, we want to proceed by reducing the amount of computation .

Let us recall the following Fraction Arithmetic :

If

$$ \begin{align}\frac {a_1}{b_1}=\frac {a_2}{b_2}=\dots=\frac {a_n}{b_n}=k\end{align} $$

with $\thinspace b_1+b_2+\dots+b_n≠0\thinspace $, then :

$$\frac {a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}=k\thinspace .$$

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Thus, by appying the Fraction Arithmetic, you have :

$$ \begin{align}\frac ad&=\frac ab\cdot\frac bc\cdot \frac cd\\ &=\frac {a^3}{b^3}=\frac {b^3}{c^3}=\frac {c^3}{d^3}\\ &=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\thinspace .\end{align} $$

which completes the proof .

Answer 3

In this answer, we generalize the problem statement and provide a straightforward solution .

Let

$$ \begin{align}\frac {a_1}{a_2}=\frac {a_2}{a_3}=\frac {a_3}{a_4}&=\thinspace\dots\thinspace=\frac {a_n}{a_{n+1}}\end{align} $$

Then we want to show that ,

$$ \begin{align}\frac {a_1}{a_{n+1}}=\frac {a_1^n+a_2^n+\thinspace \dots\thinspace+a_n^n}{a_2^n+a_3^n+\thinspace \dots\thinspace+a_{n+1}^n}\thinspace \end{align} $$

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By applying Fraction Arithmetic ,

If

$$ \begin{align}\frac {a_1}{b_1}&=\frac {a_2}{b_2}=\dots=\frac {a_n}{b_n}=k\end{align} $$

with $\thinspace b_1+b_2+\dots+b_n≠0\thinspace $, then :

$$\frac {a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}=k\thinspace .$$

we have :

$$ \begin{align}\frac {a_1}{a_{n+1}}&=\frac {a_1}{a_2}\cdot\frac {a_2}{a_3}\cdot\frac {a_3}{a_4}\cdot\thinspace\dots\thinspace\cdot\frac {a_n}{a_{n+1}}\\ &=\frac {a_1^n}{a_2^n}=\frac {a_2^n}{a_3^n}=\thinspace\dots\thinspace=\frac {a_n^n}{a_{n+1}^n}\\ &=\frac {a_1^n+a_2^n+\thinspace \dots\thinspace+a_n^n}{a_2^n+a_3^n+\thinspace \dots\thinspace+a_{n+1}^n}\end{align} $$

Thus, we obtain :

$$ \begin{align}\frac {a_1}{a_{n+1}}&=\frac {a_1^n+a_2^n+\thinspace \dots\thinspace+a_n^n}{a_2^n+a_3^n+\thinspace \dots\thinspace+a_{n+1}^n}\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

Finally note that, if $n$ is even, then $a_i≠-a_{i+1}$, where $1≤i≤n+1\thinspace .$ Because, the last equality holds, only if when $a_2^n+a_3^n+\thinspace \dots\thinspace+a_{n+1}^n≠0\thinspace .$