MHB Prove AM-GM Inequality: What Values to Use?

[delc1]

Could someone please help me with this question:

View attachment 2476

What values are we meant to use to prove this inequality?

many thanks

 

[Nono713]

Try and put things into a form where you can use the AM-GM inequality. Here you have a power of three on the LHS, which means you're probably going to have to consider AM-GM with three terms. So we try this (note the inequality sign doesn't change because the cube root function is strictly increasing over the positive reals):

$$4x + 10y \geq \sqrt[3]{1080 x^2 y}$$

But we only have two variables here! But look: there's an $x^2$ term on the RHS, and we know that here the RHS will be the geometric mean (and the LHS the arithmetic mean), so that means we probably have two terms in $x$ and one in $y$. First let's put the LHS in the right form as an actual mean, so we get:

$$\frac{12x + 30y}{3} \geq \sqrt[3]{1080 x^2 y}$$

That's looking a bit closer to what we need. So now we want to write the above in the form below where we can directly apply AM-GM:

$$\frac{ax + bx + cy}{3} \geq \sqrt[3]{abc x^2 y}$$

Can you see why once we find good values of $a$, $b$, $c$ we will be able to apply AM-GM directly? (hint: $abcx^2y = (ax)(bx)(cy)$). Can you now solve for $a$, $b$ and $c$ and conclude? (hint: $c$ is already given, so you have a system of two equations in two variables).

 

[anemone]

Hi delc1,

First I played with the sum of $4x$ and $10y$ and applied the AM-GM and I ended up with $\dfrac{4x+10y}{2}\ge (4x\cdot10y)^{\dfrac{1}{2}}_{\phantom{i}}$. This gave us the idea that if we wanted the power of 3 on one side of the inequaliy sign, we must apply the AM-GM inequality to a total of 3 terms.

Note that $1080=2^33^35$ and that the RHS of the inequality$(4x+10y)^3\ge1080x^2y$ consists of two $x$ and and one $y$, I know I must separate the $4x$ into the sum of $2x+2x$ and by applying the AM-GM to the terms $2x,\,2x,\,10y$, I get:

$\dfrac{2x+2x+10y}{3}\ge (2x\cdot2x\cdot10y)^{\dfrac{1}{3}}_{\phantom{i}}$, or equivalently,

$(4x+10y)^3\ge1080x^2y$

 

[delc1]

​

Hi delc1,

First I played with the sum of $4x$ and $10y$ and applied the AM-GM and I ended up with $\dfrac{4x+10y}{2}\ge (4x\cdot10y)^{\dfrac{1}{2}}_{\phantom{i}}$. This gave us the idea that if we wanted the power of 3 on one side of the inequaliy sign, we must apply the AM-GM inequality to a total of 3 terms.

Note that $1080=2^33^35$ and that the RHS of the inequality$(4x+10y)^3\ge1080x^2y$ consists of two $x$ and and one $y$, I know I must separate the $4x$ into the sum of $2x+2x$ and by applying the AM-GM to the terms $2x,\,2x,\,10y$, I get:

$\dfrac{2x+2x+10y}{3}\ge (2x\cdot2x\cdot10y)^{\dfrac{1}{3}}_{\phantom{i}}$, or equivalently,

$(4x+10y)^3\ge1080x^2y$

Thanks all, the help is always appreciated!