brianlee7799 said:
If a1,a2, a3 are positive, prove that 1/3(a1+a2+a3) >= cuberoot(a1*a2*a3).
This one I'm completely clueless. Please help. Thanks!!
\(\displaystyle \frac{a_1+a_2+a_3}{3}\ \ge\ \sqrt[3]{a_1a_2a_3}\ ?\)
Let's take it that the numbers are arranged in decreasing order,
if they are not, let's do it and relabel the values so that...
\(\displaystyle a_1\ \ge\ a_2\ \ge\ a_3\)
Also, let's relabel the values as follows
\(\displaystyle A_1=\sqrt[3]{a_1},\ A_2=\sqrt[3]{a_2},\ A_3=\sqrt[3]{a_3}\)
Then..
\(\displaystyle a_1+a_2+a_3\ \ge\ 3\left(a_1a_2a_3\right)^{\frac{1}{3}}\ ?\)
\(\displaystyle a_1+a_2+a_3\ \ge\ 3\left(a_1\right)^{\frac{1}{3}}\left(a_2\right)^{\frac{1}{3}}\left(a_3\right)^{\frac{1}{3}}\ ?\)
\(\displaystyle \left(A_1\right)^3+\left(A_2\right)^3+\left(A_3\right)^3\ \ge\ 3A_1A_2A_3\ ?\)
Now here's a little trick, which explains why the numbers ought to be in decreasing order.
\(\displaystyle a\ \ge\ b\)
\(\displaystyle c\ \ge\ d\)
\(\displaystyle a-b\ \ge\ 0\)
\(\displaystyle c-d\ \ge\ 0\)
\(\displaystyle (a-b)(c-d)\ \ge\ 0\)
\(\displaystyle ac-ad-bc+bd\ \ge\ 0\)
\(\displaystyle ac+bd\ \ge\ ad+bc\)
Notice that we begin with
\(\displaystyle ac+bd=ac+bd\)
and move to
\(\displaystyle ac+bd\ \ge\ ad+bc\)
Hence, if we swop the smaller d with the larger c, we go from an equality to an inequality.
We can do the same thing in the problem given.
\(\displaystyle \left(A_1\right)^2A_1+\left(A_2\right)^2A_2+\left(A_3\right)^2A_3=\left(A_1\right)^3+\left(A_2\right)^3+\left(A_3\right)^3\)
Now swop the smaller \(\displaystyle A_3\) with the larger \(\displaystyle A_1\) to form the inequality
\(\displaystyle \left(A_1\right)^2A_3+\left(A_2\right)^2A_2+\left(A_3\right)^2A_1\ \le\ \left(A_1\right)^3+\left(A_2\right)^3+\left(A_3\right)^3\)
Swop an \(\displaystyle A_2\) from the centre with an \(\displaystyle A_3\) from the third
\(\displaystyle \left(A_1\right)^2A_3+\left(A_2\right)^2A_3+A_3A_2A_1\ \le\ \left(A_1\right)^3+\left(A_2\right)^3+\left(A_3\right)^3\)
Swop an \(\displaystyle A_1\) from the first with an \(\displaystyle A_2\) from the centre
\(\displaystyle A_1A_2A_3+A_1A_2A_3+A_1A_2A_3\ \le\ \left(A_1\right)^3+\left(A_2\right)^3+\left(A_3\right)^3\)
If you rewrite in the original form, we have
\(\displaystyle 3\sqrt[3]{a_1a_2a_3}\ \le\ a_1+a_2+a_3\)
