Prove $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $ Let $u,v,w>0$ and $a,b,c$ are positive constant. Prove that $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $
First, I prove with $x+y+z=1$ so $x^ay^bz^c\le\left(\frac{a}{a+b+c}\right)^a\left(\frac{b}{a+b+c}\right)^b\left(\frac{c}{a+b+c}\right)^c$ by Largrange theorem
And it become $\left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c\le\left(\frac{1}{a+b+c}\right)^{a+b+c}=\left(\frac{x+y+z}{a+b+c}\right)^{a+b+c}$
So it true with $x+y+z=1$ but i can't prove it true with $x,y,z>0$. Please help me! Thank you.
 A: The inequality is homogenenous in $(u, v, w)$: If you have proven
$$
\left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c\le\left(\frac{1}{a+b+c}\right)^{a+b+c}
$$
under the condition $x+y+z=1$ then you can use that to prove the general case: With
$$
 x = \frac{u}{u+v+w}, y = \frac{v}{u+v+w},z = \frac{w}{u+v+w}
$$
one has $x+y+z=1$ and
$$
\left(\frac{u}{a}\right)^a\left(\frac{v}{b}\right)^b\left(\frac{w}{c}\right)^c = \left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c (u+v+w)^{a+b+c} \\
\le \left(\frac{1}{a+b+c}\right)^{a+b+c}(u+v+w)^{a+b+c} 
= \left(\frac{u+v+w}{a+b+c}\right)^{a+b+c} \, .
$$
A: Since we have positive numbers involved therefore we can use weighted $A.M-G.M$ on $\frac{u}{a},\frac{v}{b},\frac{w}{c}$ with weights respectively $a,b,c$
$$\displaystyle\bigg({\frac{\sum_ia_ix_i}{\sum_i a_i}}\bigg)^{\sum_i a_i}\geq\bigg(\Pi (x_i)^{(a_i)}\bigg) $$
Here $x_1=\frac{u}{a} $ and $a_1=a$ ,  $x_2=\frac{v}{b} $ and $a_2=b$ and  $x_3=\frac{w}{c} $ and $a_3=c$
Equality holds when $\frac{u}{a}=\frac{v}{b}=\frac{w}{c}$
