Why is my solution incorrect? AM GM inequality

Question

The problem is: Let $p+2q=3$ and find the maximum of $pq$. Of course I could solve this as below: $$pq=(3-2q)q=-2(q-\frac{3}{4})^2+\frac{9}{8}$$ so, the maximum is $\frac{9}{8}$.

However, I got another solution i.e. using AMGM inequality. $$\sqrt{pq} \leq \frac{p+q}{2}$$ This gave me the answer $\text{max} (pq)=1$ and I think this is wrong. But I have no idea what is wrong in AMGM solution.

AMGM solution $\sqrt{pq} = \frac{p+q}{2}$ iff $p=q$. so let $p=q$ and then $p=q=1$. This means $pq\leq 1^2=1$, so the maximum of $pq$ is $1$.

I appreciate any help, thanks in advance!

Answer 1

it is about your second solution : we know $\sqrt{xy} \leq \frac{x+y}{2}$ so use it correctly.

we suppose $p+2q=3$ then take $x=p , y=2q$ $$\color{red}{\sqrt{(p)(2q)} \leq \frac{(p)+(2q)}{2}}$$ this mean $$\sqrt{2pq)} \leq \frac{3}{2}$$ to the power of two $$2pq\leq \dfrac94 \to pq \leq \dfrac98 \to \max{(pq)}=\dfrac98$$

Answer 2

AM-GM tells you that $\sqrt{pq}\leq\frac{p+q}{2}$ with equality iff $p=q$. In particular, this implies that if $p+q$ is held constant, then $pq$ is maximized when $p=q$, since $\sqrt{pq}$ must always be at most half of the constant value of $p+q$.

But in your problem, $p+q$ is not constant. So you know that $\sqrt{pq}\leq\frac{p+q}{2}$ is always true, but you can't say $p+q=2$ to conclude that $pq\leq 1$. For instance, if $p=2$ and $q=1/2$, then $p+q=5/2$, and so all AM-GM tells you is that $\sqrt{pq}<\frac{5}{4}$ so $pq< \frac{25}{16}$.

Answer 3

We have for $a,b \ge 0$,

$$\sqrt {ab}\le \frac {a+b}{2}$$

if $a=b $ then $\sqrt {ab}=\frac {a+b}{2} $

but if $a\ne b $ then $$\sqrt {ab} <\frac {a+b}{2} $$

the max of $\sqrt {ab }$ could not be $\frac {a+b}{2} $, it depends on the condition satisfied by $a $ and $b $.

For example, suppose that the condition is $$p+2q=4$$ then the max is attained for $p=2$ and $q=1$. $$\sqrt {pq}=\sqrt {2}\approx 1. 414 $$ $$\frac {p+q}{2}=\frac {3}{2}=1.5$$

Answer 4

Indeed, for every given pair of $(p,q)$, $\displaystyle\sqrt{pq}\le\frac{p+q}2$ holds. But the existance of a pair $(p_0,q_0)$ such that the equaity holds does not imply that it gives the maximum $pq$ for all pairs of $(p,q)$. There might be some pair $(p',q')$ such that $\displaystyle\sqrt{p'q'}\neq\frac{p'+q'}2$ but $p'q'>p_0q_0$.