Show that $1/ac + 1/bc \ge 16 $ if $a, b, c>0$ and $a+b+c=1$ (Seen on Quora at
https://www.quora.com/How-can-I-prove-1-ac-1-bc-16-if-a-b-c-0-and-satisfy-a-b-c-1)
Show that
$1/ac + 1/bc \ge 16
$ if
$a, b, c>0$ and
$a+b+c=1$.
Here is my answer.
I would be interested in any others.
$1/ac + 1/bc \ge 16\\
\iff b+a \ge 16abc\\
\iff 1-c \ge 16abc\\
\iff 1 \ge c(1+16ab)\\
$
If $a=b$,
$1=2a+c$
so
$a=b
=(1-c)/2
$
so
$ab=(1-c)^2/4
$
so
$\begin{array}\\
c(1+16ab)-1
&=c(1+4(1-c)^2)-1\\
&=4c^3-8c^2+5c-1\\
&=(2c-1)^2(c-1)\\
&\le 0
\qquad\text{for } 0 \le c \le 1\\
\end{array}
$
So this is true when $a=b$.
If we can show that
$\dfrac1{a}+\dfrac1{b}
\ge 2\dfrac1{(a+b)/2}$,
we are done since
$2\dfrac1{(a+b)/2}
\ge 16c
$.
This is
$\dfrac1{a}+\dfrac1{b}
\ge 4\dfrac1{a+b}
$.
$\begin{array}\\
\dfrac1{a}+\dfrac1{b}- 4\dfrac1{a+b}
&=\dfrac{b(a+b)+a(a+b)-4ab}{ab(a+b)}\\
&=\dfrac{(a+b)^2-4ab}{ab(a+b)}\\
&=\dfrac{(a-b)^2}{ab(a+b)}\\
&\ge 0\\
\end{array}
$
and we are done.

Another way (basically, we reduce the given inequality to one variable inequality):
$$
\frac{1}{ac}+\frac{1}{bc}=\frac{a+b}{abc}\ge\frac{4(a+b)}{(a+b)^2c}=\frac{4}{c(1-c)}\ge16.
$$
Comment. As @DanielWainfleet pointed out in the comments, this approach shows that we get equality only if $a=b$ (the first inequality) and $c=1-c$ (the second one), i.e. when $(a,b,c)=(1/4,1/4,1/2)$.
