Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x,y,z,b,c,d \in \mathbb{R}$ with the properties $x \geq 0$, $x+y \geq 0$, $x+y+z \geq 0$, $b \geq c \geq d >1$. Prove that for any $a >1$ the following inequalities:

a) $${b^{a^x}}\cdot {c^{a^y}}\geq bc.$$

b)$${b^{a^x}}\cdot {c^{a^y} } \cdot {d^{a^{z}}}\geq bcd.$$

I'm not able to solve this inequality using $AM\geq GM$. I appreciate your help. Thanks :)

share|cite|improve this question
up vote 5 down vote accepted

Taking the logarithm of both sides, we are trying to prove the inequality $$\left(a^{x}-1\right)\log b+\left(a^{y}-1\right)\log c+\left(a^{z}-1\right)\log d\geq0.$$ For positive $a>1$, we have that $$a^{x}-1\geq x\log a,$$ and so $$\left(a^{x}-1\right)\log b+\left(a^{y}-1\right)\log c+\left(a^{z}-1\right)\log d\geq\log a\left(x\log b+y\log c+z\log d\right).$$ Since $x\geq0,$ and $\log b\geq\log c$, this is $$\geq\log a\left((x+y)\log c+z\log d\right).$$ Since $x+y\geq0 $ and $\log c\geq\log d$ , this is $$\geq\log a\log d\left(x+y+z\right)\geq0,$$ where the last inequality follows since $x+y+z\geq 0$ and $\log a,\log d\geq 0.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.