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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!

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closed as off-topic by Saad, José Carlos Santos, Chris Custer, choco_addicted, A. Goodier Apr 9 '18 at 9:00

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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.

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  • $\begingroup$ Yes. The case $c=a$ is obvious. $\endgroup$ – Michael Rozenberg Apr 8 '18 at 10:23