Prove the product is less than or equal to 1

[anemone]

Let $a,\,b,\,c$ be real numbers greater than $2$ such that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$.

Prove that $(a-2)(b-2)(c-2)\le 1$.

 

[MarkFL]

My solution:

Spoiler

Let:

\(\displaystyle f(a,b,c)=(a-2)(b-2)(c-2)\)

Because of cyclic symmetry, we know the extremum of the objective function will occur for:

\(\displaystyle a=b=c=3\)

and we see that:

\(\displaystyle f(3,3,3)=1\)

If we pick another point on the curve satisfying the constraint, we find:

\(\displaystyle f\left(\frac{5}{2},\frac{5}{2},5\right)=\frac{3}{4}<1\)

Hence, we may conclude:

\(\displaystyle f_{\max}=1\)

 

[anemone]

Well done MarkFL! And thanks for participating in my challenge!

My solution:

Spoiler

Note that

$\begin{align*}(a−2)(b−2)(c−2)&=abc\left(\dfrac{a-2}{a}\right)\left(\dfrac{b-2}{b}\right)\left(\dfrac{c-2}{c}\right)\\&=abc\left(1-\dfrac{2}{a}\right)\left(1-\dfrac{2}{b}\right)\left(1-\dfrac{2}{c}\right)\\&=abc\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{2}{a}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{2}{b}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{2}{c}\right)\\&=abc\left(\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a}\right)\left(\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)\end{align*}$

We now use the famous identity that says for all real and positive $x,\,y$ and $z$, we have

$xyz\ge (x+y-z)(x+z-y)(y+z-x)$

In our case, we have $x=\dfrac{1}{a},\,y=\dfrac{1}{b},\,z=\dfrac{1}{c}$ and so we get

$\dfrac{1}{abc}\ge \left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)\left(\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}\right)\left(\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a}\right)$

i.e.

$abc\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)\left(\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}\right)\left(\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a}\right)\le 1$

The proof is then follows.

Equality occurs when $a=b=c=3$.