<s>[INST] 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.
 [/INST] Here's another way using  GM-HM and then AM-GM:
$$\frac{ 2}{ \frac{1}{ac} + \frac{1}{bc} } \leq  \sqrt{ abc^2 } \leq \sqrt{4} \left( \frac{  a + b + c/2 + c /2 } {4 }\right)^2 = \frac{1}{8}.$$
Equality holds when $ a = b = c/2 = \frac{1}{4}$.
If helpful to visualize the middle inequality, it is obtained by squaring $ \sqrt[4]{ ab \frac{c^2}{4} } \leq \frac{ a + b + c/2 + c/2 } { 4}$, and multiplying by the denominator.
</s>