question:
maximise the following function $f$
$$f= x^p y^q z^r$$ subject to the constraint
$$ax+by+cz=p+q+r$$
i know how to do it using lagrange method of multipliers .
but i'm looking for an answer which uses simple inequalities to maximise this function
my attempt: i also tried solving it using inequality but didn't get right answer
since i've to maximise product it will be maximum when all three terms in sum are equal i.e, $ax=by=cz= \dfrac{p+q+r}{3}$ then, i plug values of$ x,y,z $in terms of $a,b,c,p,q,r$ in above function$ f $ to get
$f_{max}=\dfrac{(p+q+r)^{p+q+r}}{3^{p+q+r}a^p . b^q.c^r}$
but answer is coming $\left(\dfrac{p}{a}\right)^p$ $.\left(\dfrac{q}{b}\right)^q$$\left(\dfrac{r}{c}\right)^r$
i don't know where i did wrong
please point my mistake and give plausible answer. thank you
Because by Jensen for $\ln$ we obtain: $$\ln{x^py^qz^r}=p\ln{x}+q\ln{y}+r\ln{z}=$$ $$=p\ln{p}+q\ln{q}+r\ln{r}-p\ln{a}-q\ln{b}-r\ln{c}+p\ln\frac{ax}{p}+q\ln\frac{by}{q}+r\ln\frac{cz}{r}=$$ $$=\ln{\frac{p^pq^qr^r}{a^pb^qc^r}}+(p+q+r)\left(\frac{p}{p+q+r}\ln\frac{ax}{p}+\frac{q}{p+q+r}\ln\frac{by}{q}+\frac{r}{p+q+r}\ln\frac{cz}{r}\right)\leq$$ $$\leq\ln{\frac{p^pq^qr^r}{a^pb^qc^r}}+(p+q+r)\ln\frac{ax+by+cz}{p+q+r}=\ln{\frac{p^pq^qr^r}{a^pb^qc^r}}.$$ Easy to see that the equality occurs, which says that we got a maximal value: $$\left(\dfrac{p}{a}\right)^p\left(\dfrac{q}{b}\right)^q\left(\dfrac{r}{c}\right)^r$$