Inequalities involving QM-AM-GM-HM or Cauchy Schwarz I was wondering how to do an inequality problem involving QM-AM-GM-HM.
Question: For positive $a$, $b$, $c$ such that $\frac{a}{2}+b+2c=3$, find the maximum of $\min\left\{ \frac{1}{2}ab, ac, 2bc \right\}$.
I was thinking maybe apply AM-GM, however, I'm not sure what to plug in. Any help would be appreciated, thanks!
 A: Applying AM-GM
$$1=\frac{(a/2)+b+2c}{3}\ge \sqrt[3]{(a/2)b(2c)}=\sqrt[3]{abc}$$
We square both sides to get $$\sqrt[3]{a^2b^2c^2}\le 1 ~~~~~~(1)$$
Now suppose $\min\{ab/2,ac,2bc\}=k$.  We take the geometric mean of these three: 
$$\sqrt[3]{a^2b^2c^2}=\sqrt[3]{(ab/2)ac(2bc)}\ge \sqrt[3]{k^3}=k ~~~~~~~(2)$$
Combining (1) and (2), we get $k\le 1$. In fact $k=1$ can be achieved, by taking $a=2, b=1, c=0.5$.  Hence the desired answer is $1$.
A: Another way:  If possible, let the optimum occur when one among $\frac12ab, ac, 2bc$ is lesser than the others.  Then note that this smaller term determines the maximum and can be increased at the expense of the variable which is not involved in it.  Thus at optimum we must have $\frac12ab = ac = 2bc$, which with the constraint gives $\frac12a=b=2c=1$ as the only nonzero solution and leads to a maximum of $1$.
A: Hint:
$$\frac{\frac{a}{2}+b}{2}\ge\sqrt{\frac{ab}{2}}\iff \left(\frac{\frac{a}{2}+b}{2}\right)^2\ge\frac{ab}{2}$$
$$\frac{2c+b}{2}\ge\sqrt{2bc}\iff \left(\frac{2c+b}{2}\right)^2 \ge 2bc$$
$$\frac{\frac{a}{2}+2c}{2}\ge\sqrt{ac}\iff \left(\frac{\frac{a}{2}+2c}{2}\right)^2\ge ac$$
Note: 
 $$\frac{x+y}{2}\ge\sqrt{xy}$$
Equality occurs when  $x=y$.
