I want to prove:
For $a, b, c>0$ and $a\leq b\leq c$ then $b^{c- a}\geq a^{c- b}c^{b- a}$
From this inequality, I can prove the under inequality easily: $$a_{1}x_{1}+ a_{2}x_{2}\geq x_{1}^{a_{1}}+ x_{2}^{a_{2}}$$ with $x_{1}+ x_{2}= 1$
I try to use logarithm two both sides but I can' t
Thus, I need to the help, thanks!
We will examine our inequality by take the logarithm both sides: $$\log b\ge \frac{c-b}{c-a}\log a + \frac{b-a}{c-a}\log c. $$
Now observe that $$\frac{c-b}{c-a} a + \frac{b-a}{c-a} c = b.$$ Therefore, as logarithm is concave, our inequality directly follows.