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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 :)

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1 Answer 1

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 $\log a,\log d\geq 0.$

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