# Inequality $\frac1a + \frac1b + \frac1c \leq \frac{a^8+b^8+c^8}{a^3b^3c^3}$

Let $$a,b,c$$ be positive reals .

Prove that $$\displaystyle \frac1a + \frac1b + \frac1c \leq \frac{a^8+b^8+c^8}{a^3b^3c^3}$$

I found this one in a book, no hints mentioned, but marked as very hard.

I can't make any progress...

• Did you get this problem from a book discussing algebraic inequalities? Commented Jun 3, 2015 at 19:09
• @user74973 No, in a book about real analysis. Commented Jun 4, 2015 at 5:18
• What is the title of the real analysis book? Do you think that author intended an argument using the Arithmetic-Geometric Mean Inequality? Commented Jun 4, 2015 at 14:57

Hint: Use AM-GM inequality we have $$3a^8+3b^8+2c^8\ge 8a^3b^3c^2$$ proof: since $$3a^8+3b^8+2c^8=a^8+a^8+a^8+b^8+b^8+b^8+c^8+c^8\ge 8(a^{8+8+8}b^{8+8+8}c^{8+8})^{1/8}=8a^3b^3c^2$$ similar $$3a^8+2b^8+3c^8\ge 8a^3b^2c^3$$ $$2a^8+3b^3+3c^3\ge 8a^2b^3c^3$$ so $$8(a^8+b^8+c^8)\ge 8a^2b^2c^2(ab+bc+ac)$$ so $$\dfrac{a^8+b^8+c^8}{a^3b^3c^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$
We might make repeated use of the inequality $x^2+y^2+z^2 \ge xy+yz+zx$, for $x,y,z \in \mathbb{R}$ as well.
We have infact: $$\displaystyle a^8+b^8+c^8 \ge \sum\limits_{cyc} a^4b^4 \ge \sum\limits_{cyc} a^2b^4c^2 \ge \sum\limits_{cyc} ab^2c.bc^2a = \sum\limits_{cyc} a^2b^3c^3 = a^3b^3c^3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$$