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PROBLEM

If a, b, c, d are four distinct positive quantities in G.P., then show that a + d > b + c

Solution in the book:

A.M. > G.M. for the first three terms

$(a+c)>2b$; since $ac=b^2$ ................(A)

Similarly, for the last three terms

$(b+d)>2c$; since $bd=c^2$ ................(B)

Adding (A) and (B), we get (a + c) + (b + d) > 2b + 2c

a + d > b + c

Solution done by me:

where a is 1st term and r is common ratio

a=a

b=ar

c=ar^2

d=ar^3

A.M. > G.M. for the first and fourth terms

$(a+d)/2$ > root of ad

$(a+d)/2$ > 2ar root of r; ................(A)

Similarly, for the second and third terms

$(b+c)/2$ > root of bc

$(b+c)/2$ > 2ar root of r; ................(B)

(A) = (B), we get

a + d = b + c

Where have i went wrong??

Marble
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    Your mistake lies in the conclusion that $A = B$. It is tantamount to saying that if $x > k$ and $y > k$ then $x = y$, which is obviously not true. – jgsmath Jul 22 '17 at 15:58
  • but am>=gm then it is valid – Marble Jul 22 '17 at 16:00
  • $A$ holds because $AM \geq GM$. $B$ holds because $AM \geq GM$. But that does not mean that LHS of $A$ and LHS of $B$ are equal. – jgsmath Jul 22 '17 at 16:02
  • Also, I'd suggest that you use Mathjax. The text will be much more readable. – jgsmath Jul 22 '17 at 16:06
  • Note that in the book solution AM>GM, because the equality holds if $a=c$ in $a+c\ge 2\sqrt{ac}$, which is excluded by the condition "distinct quantities". – farruhota Jul 22 '17 at 16:55

2 Answers2

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Using AM-GM seems quite convoluted here.

Let $a=a, b = ar, c = ar^2, d = ar^3$.

$a,b,c,d$ are distinct positive $\implies$ $r$ is positive and not equal to 1.

Given inequality then boils down to proving $1 + r^3 > r(1+r)$, which is equivalent to proving $1 - r + r^2 > r \iff (1-r)^2 > 0$ which is obviously true. Strict inequality holds because $r \neq 1$.

jgsmath
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WLOG the four terms are $$A/D^3, A/D, AD, AD^3$$ where $D>0,\ne1$

$$a+d=A\left(D^3+\dfrac1{D^3}\right)$$

$$b+c=A\left(D+\dfrac1D\right)$$

$$a+d-(b+c)=A\left(D+\dfrac1D\right)\left(D^2-2+\dfrac1{D^2}\right)==A\left(D+\dfrac1D\right)\left(D-\dfrac1D\right)^2$$

lab bhattacharjee
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