Vriska said:
This really doesn't help , this question was asked in a regional math olympiad to 10th graders so this information doesn't mean much to me . I need a simple proof using the AM GM theorem and also maybe a way to prove this inequality works for x and y lesser than 2 but greater than 0
Here is an elementary argument that comes close, but does not quite do what you want. First, note that if ##t_1, t_2, \ldots, t_6 > 0## are some positive numbers and ##a_1, a_2, \ldots, a_6 > 0, \; a_1 + a_2 + \cdots + a_6 = 1## are positive "weights" summing to 1, then
a_1 t_1 + a_2 t_2 + \cdots + a_6 t_6 \geq t_1^{a_1} \cdot t_2^{a_2} \cdots t_6^{a_6}
This is a generalization of the AM/GM inequality, Apply it when the left-hand-side is written as
\text{LHS} = \frac{1}{6} (24 x^4) + \frac{1}{4} (16 y^3) + \frac{1}{6} (30 x^2) + \frac{1}{4} (4y) + \frac{1}{6} (6)
That will give you
\text{LHS} \geq (24 \cdot 30 \cdot 6)^{1/6} \, (16 \cdot 4)^{1/4} xy = 4320^{1/6} 64^{1/4} xy \doteq 11.41455411\, xy
This gives LHS >= 11.416 xy instead of LHS >= 12 xy. So, as I said: close, but not quite what you want. Again, I don't know if this helps you.
Note added in edit: perhaps it is possible to group the terms on the left in another way and with different weights, allowing use of the more general AM/GM inequality
\sum_{i=1}^m a_i t_i \geq \prod_{i=1}^m t_i^{a_i}
where ##t_i, a_i > 0, i=1, \ldots m## and ##\sum a_i = 1##.
Edit note 2: we can get a constant >= 12 on the right by looking at the problem as follows: if
t_1 = 4x^2, \: t_2 = 4 y^3, \: t_3 = 5x^2, \: t_4 = y, \: t_5 = 1
we have
f(x,y) = t_1 + t_2 + t_3 + t_4 + t_5 \\<br />
= a_1(t_1/a_1) + a_2 (t_2/a_2) + a_3 (t_3/a_3) + a_4 (t_4/a_4) + a_5 (t_5/a_5 )\\<br />
\geq (t_1/a_1)^{a_1} (t_2/a_2)^{a_2} (t_3/a_3)^{a_3} (t_4/a_4) ^{a_4} (t_5/a_5)^{a_5}\\<br />
= (4/a_1)^{a_1}(4/a_2)^{a_2} (5/a_3)^{a_3} (1/a_4)^{a_4} (1/a_5)_{a_5} x^{4a_1+ 2a_3} y^{3a_2+a_4}
Thus, ##f(x,y) \geq c xy##, if
4 a_1 + 2 a_3 = 1,\\<br />
3 a_2 + a_4 = 1\\<br />
a_1 + a_2 + a_3 + a_4 + a_5 = 1
For given ##a_i## the constant ##c## on the right is
c = \left( \frac{4}{a_1}\right)^{a_1} \left( \frac{4}{a_2} \right)^{a_2}<br />
\left( \frac{5}{a_3} \right)^{a_3} \left( \frac{1}{a_4} \right)^{a_4} \left( \frac{1}{a_5} \right)^{a_5}
There are many values of the ##a_i## that give ##c \geq 12##. (In particular, if we maximize ##c## subject to the constraints on the ##a_i## we can get ##c = 14.4902646505479994 > 12##. (In other words, we actually have ##f(x,y) \geq 14.49 xy##.) If you do not care to use an optimization package (or are not allowed to do so) you can just try some values manually until you stumble upon a value ##c \geq 12##.
It would help to use the restrictions on the ##a_i## to solve for three of them in terms of the other two; for example, we can easily solve for ##a_3, a_4, a_5## in terms of ##a_1,a_2##. Then ##c## can be re-written in terms of ##a_1, a_2## only, and there are some restrictions on ##a_1, a_2## due to the requirements ##a_3,a_4,a_5 \geq 0##. One can then just try some trial values of ##(a_1,a_2)## until we get a value ##c \geq 12##. In principle, all you need is a good hand-held calculator.