Prove:if 1/a+1/b+1/c=1 and a,b,c >0 then (a-1)(b-1)(c-1)>=8

[Danijel]

The problem: prove that if 1/a + 1/b + 1/c=1 and a,b,c are positive numbers then (a-1)(b-1)(c-1)>=8

I've tried it myself but couldn't do it. I have tried going backwards i.e. :
(a-1)(b-1)(c-1)>=8
(fast forward) abc -(ab+bc+ac) +a+b+c -1 >=8
Now from 1/a + 1/b + 1/c = 1 we have that ab+bc+ac/abc=1 thus :
ab + bc + ac = abc , so abc -(ab+bc+ac) +a+b+c -1>=8 is really a+b+c -1>=8
that is a+b+c >=9 . So If I could somehow prove that a+b+c>9 then I could work backwards and prove the former statement.
But the only progress I've made is :all of a,b,c must be greater than 1 because if they weren't then any of the 1/a,1/b,1/c would be greater then 1 thus we would need negatives(and a,b,c are positive) to "bring back" the value to 1, nor can they equal to 1 because the same thing would happen(one of 1/a,1/b,1/c would be equal to 1 and we would need negatives)so a>1, b>1, c>1 adding these three together we get a+b+c >3 ,and I need a+b+c+>=9.
Sorry for the long post. Could anyone give me some directions?

Thanks,
Daniel

 

[RUber]

Could you prove it for 2?
$\frac1a + \frac1b = 1$ Prove that $ab\geq 4$?

 

[RUber]

Your problem hinges on the relationship between a, b, and c.
If a = b = c, then they are all 3, right?
Let's change one of the three...let a be (3-x), either b or c would have to increase to maintain the 1/a + 1/b + 1/c = 1 balance.
How much would it change?
Let's say they absorb the difference equally...share the burden, then you have $\frac{1}{3-x} + \frac{2}{3+y} = 1$. Can you solve for y in terms of x? If y > x/2 you have shown that (3-x) + (3+y)+(3+y) > 9.

 

[Danijel]

Thank you RUber!

 

[Ray Vickson]

The problem: prove that if 1/a + 1/b + 1/c=1 and a,b,c are positive numbers then (a-1)(b-1)(c-1)>=8

I've tried it myself but couldn't do it. I have tried going backwards i.e. :
(a-1)(b-1)(c-1)>=8
(fast forward) abc -(ab+bc+ac) +a+b+c -1 >=8
Now from 1/a + 1/b + 1/c = 1 we have that ab+bc+ac/abc=1 thus :
ab + bc + ac = abc , so abc -(ab+bc+ac) +a+b+c -1>=8 is really a+b+c -1>=8
that is a+b+c >=9 . So If I could somehow prove that a+b+c>9 then I could work backwards and prove the former statement.
But the only progress I've made is :all of a,b,c must be greater than 1 because if they weren't then any of the 1/a,1/b,1/c would be greater then 1 thus we would need negatives(and a,b,c are positive) to "bring back" the value to 1, nor can they equal to 1 because the same thing would happen(one of 1/a,1/b,1/c would be equal to 1 and we would need negatives)so a>1, b>1, c>1 adding these three together we get a+b+c >3 ,and I need a+b+c+>=9.
Sorry for the long post. Could anyone give me some directions?

Thanks,
Daniel

You can regard this as a constrained optimization problem:
\begin{array}{l} \min_{a,b,c} f(a,b,c) = (a-1)(b-1)(c-1) \\<br /> \text{subject to} \\<br /> \displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1<br /> \end{array}<br />
Let us restrict the problem to $a,b,c > 1$. There is a good reason for this, because if you allow any of $a,b,c$ to be $< 1$ but positive, then you must also allow one or more of them to be $< 0$ in order to balance out the constraint. In that case you can construct a sequence of feasible $a_n, b_n, c_n$ values that give $f(a_n,b_n,c_n) \to -\infty$: an unbounded problem having no minimum.

OK, so we want $a, b, c > 1$. Solve the constraint for $c$ as a function of $a,b$, then plug that formula into $f(a,b,c)$ to get a new function $F(a,b)$ involving $a$ and $b$ alone, and with no constraint anymore (other than $a,b > 1$, and implicitly, $c > 1$). This unconstrained 2-variable optimization problem can be solved by finding the stationary point (or points) of $F(a,b)$. If $\min_{a,b} F(a,b) \geq 8$ you are done.

 

[Danijel]

You can regard this as a constrained optimization problem:
\begin{array}{l} \min_{a,b,c} f(a,b,c) = (a-1)(b-1)(c-1) \\<br /> \text{subject to} \\<br /> \displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1<br /> \end{array}<br />
Let us restrict the problem to $a,b,c > 1$. There is a good reason for this, because if you allow any of $a,b,c$ to be $< 1$ but positive, then you must also allow one or more of them to be $< 0$ in order to balance out the constraint. In that case you can construct a sequence of feasible $a_n, b_n, c_n$ values that give $f(a_n,b_n,c_n) \to -\infty$: an unbounded problem having no minimum.

OK, so we want $a, b, c > 1$. Solve the constraint for $c$ as a function of $a,b$, then plug that formula into $f(a,b,c)$ to get a new function $F(a,b)$ involving $a$ and $b$ alone, and with no constraint anymore (other than $a,b > 1$, and implicitly, $c > 1$). This unconstrained 2-variable optimization problem can be solved by finding the stationary point (or points) of $F(a,b)$. If $\min_{a,b} F(a,b) \geq 8$ you are done.

Wow, I would've never thought of something like that. Although this way of proving is a little too complex for me.
Thank you for your reply!