Question

If a+b/2,a+c/2 and b+c/2 then find minimum value of a+b+c

Answer 1

19

Answer 2

Solution:

We are given that

$$\frac{a+b}{2},\quad \frac{a+c}{2},\quad \frac{b+c}{2}$$

form a geometric progression. Thus, the middle term squared equals the product of the extreme terms:

$$\left(\frac{a+c}{2}\right)^2=\frac{a+b}{2}\cdot\frac{b+c}{2}.$$

Multiplying by 4, we have:

$$(a+c)^2 = (a+b)(b+c).$$

A useful method is to write:

$$\frac{a+b}{2}=x,\quad \frac{a+c}{2}=xr,\quad \frac{b+c}{2}=xr^2,$$

with $r>1$ (since $a<b<c$). Therefore:

$$a+b=2x,\quad a+c=2xr,\quad b+c=2xr^2.$$

Express $a$, $b$, and $c$ in terms of $x$ and $r$. Adding $a+b$ and $a+c$ gives:

$$2a+b+c=2x+2xr.$$

But $b+c=2xr^2$, so:

$$2a=2x+2xr-2xr^2\quad \Longrightarrow\quad a=x(1+r-r^2).$$

Then, from $a+b=2x$:

$$b=2x-a=2x-x(1+r-r^2)=x(1-r+r^2).$$

Similarly, from $a+c=2xr$:

$$c=2xr-a=x(2r-1-r+r^2)=x(r+r^2-1).$$

To make $a$, $b$, and $c$ positive integers, choose $r=\frac{3}{2}$. Then:

$$a=x\left(1+\frac{3}{2}-\frac{9}{4}\right)=x\left(\frac{4+6-9}{4}\right)=\frac{x}{4},$$ $$b=x\left(1-\frac{3}{2}+\frac{9}{4}\right)=x\left(\frac{4-6+9}{4}\right)=\frac{7x}{4},$$ $$c=x\left(\frac{3}{2}+\frac{9}{4}-1\right)=x\left(\frac{6+9-4}{4}\right)=\frac{11x}{4}.$$

Choose the smallest $x$ allowing integers, namely $x=4$. Then:

$$a=1,\quad b=7,\quad c=11.$$

Thus:

$$a+b+c=1+7+11=19.$$

Minimal Explanation:

1. For GP: $(a+c)^2 = (a+b)(b+c)$.
2. Let $\frac{a+b}{2}=x$, $\frac{a+c}{2}=xr$, $\frac{b+c}{2}=xr^2$. Solve to get: $$a=x(1+r-r^2),\quad b=x(1-r+r^2),\quad c=x(r+r^2-1).$$
3. With $r=\frac{3}{2}$ and $x=4$, obtain $a=1$, $b=7$, $c=11$ so $a+b+c=19$.