Prove that $(a_{1}b_{1}+a_{2}b_{2})^{a_{1}+a_{2}}>b_{1}^{a_{1}}b_{2}^{a_{2}}(a_{1}+a_{2})^{a_{1}+a_{2}}$ I have problems to prove the following inequality:
Let $a_{1}$, $a_{2}$, $b_{1}$, $b_{2}$ $\in$ $\mathbb{R}^{+}$ then:
$$(a_{1}b_{1}+a_{2}b_{2})^{a_{1}+a_{2}}>b_{1}^{a_{1}}b_{2}^{a_{2}}(a_{1}+a_{2})^{a_{1}+a_{2}}$$
where: $a_{1}$, $a_{2}$, $b_{1}$, $b_{2}$ are all different.
I was unable to make the show, and I begin to doubt that is true why come to ask if it is valid or not, should be true if I can help with testing, I was the grateful
 A: Basically the same proof in a different disguise based on Jensen's inequality, the mother of many other inequalities: by rearranging into 
$$
\left(\frac{a_1b_1+a_2b_2}{a_1+a_2}\right)^{a_1+a_2}>b_1^{a_1}b_2^{a_2},
$$ 
taking the natural logarithm both sides (an increasing function), and keeping the notation from the answer above, we get
$$
\ln(\omega_1b_1+\omega_2b_2)>\omega_1\ln b_1+\omega_2\ln b_2.
$$
This is Jensen's inequality for the concave-down function $\ln$ and the weights $\omega_1$, $\omega_2$, $\omega_1+\omega_2=1$. Inequality is strict if $\omega_1$ and $\omega_2$ are not zero.
A: Hint: Let $\omega_1 = \frac{a_1}{a_1+a_2}$ and $\omega_2 = \frac{a_2}{a_1+a_2}$, $\omega_1+\omega_2=1$. The inequality can be re-written as (assuming positivity of all numbers):
$$ \left(\frac{b_1}{\omega_1b_1+\omega_2b_2}\right)^{\omega_1}\left(\frac{b_2}{\omega_1b_1+\omega_2b_2}\right)^{\omega_2}<1.$$
This, in turn, using $x_1 = \frac{b_1}{\omega_1b_1+\omega_2b_2}$ and $x_2 = \frac{b_2}{\omega_1b_1+\omega_2b_2}$, is equivalent to:
$$ x_1^{\omega_1}x_2^{\omega_2}<1,$$
where $\omega_1x_1 + \omega_2x_2 = 1$, $0<x_1<1$, $0<x_2<1$, $0<\omega_1<1$, $0<\omega_2<1$, $\omega_1+\omega_2=1$. Finally, geometric mean must be less than arithmetic mean (for any system of positive weights - equality can hold in specific cases, which can be eliminated using the assumption of all numbers being positive and different). 
