If $a, b, c$ and $d$ are four distinct positive numbers in G.P, then prove that $a + d \geq b + c$.

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> [!TLDR] Write $b,c,d$ in terms of $a$ and the common ratio $r$, then show $a+d-(b+c)=a(1-r)^2(1+r)\geq0$.
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> [!FORMULA] Four consecutive G.P. terms: $a,\ b=ar,\ c=ar^2,\ d=ar^3$, where $r$ = common ratio $(r>0$ since all terms are positive$)$.

1. Since $a,b,c,d$ are in G.P. with ratio $r$: $b=ar$, $c=ar^2$, $d=ar^3$, and $a>0$, $r>0$.
2. Consider $a+d-(b+c)=a+ar^3-ar-ar^2=a\left(1+r^3-r-r^2\right)$.
3. Factor $1+r^3-r-r^2$: group as $(1-r)+(r^3-r^2)=(1-r)+r^2(r-1)=(1-r)-r^2(1-r)=(1-r)(1-r^2)$.
4. Further factor $1-r^2=(1-r)(1+r)$, so $1+r^3-r-r^2=(1-r)^2(1+r)$.
5. Hence $a+d-(b+c)=a(1-r)^2(1+r)$. …

Q.If $a, b, c$ and $d$ are four distinct positive numbers in G.P, then prove that $a + d \geq b + c$ .

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A Concrete Example

Q.If a,b,ca, b, ca,b,c and ddd are four distinct positive numbers in G.P, then prove that a+d≥b+ca + d \geq b + ca+d≥b+c.

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Q.If $a, b, c$ and $d$ are four distinct positive numbers in G.P, then prove that $a + d \geq b + c$ .

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