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orl wrote: Prove that for arbitrary positive numbers the following inequality holds \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{a^8 + b^8 + c^8}{a^3b^3c^3}.\] Multiplying this inequality with $a^3b^3c^3$, we get $a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3 \leq a^8+b^8+c^8$. This is trivial using the Muirhead inequality (in fact, $\left(8;\;0;\;0\right)\succ\left(3;\;3;\;2\right)$). Without the Muirhead inequality, one can prove this by noting that AM-GM yields $a^2b^3c^3 = aabbbccc = \sqrt[8]{a^8a^8b^8b^8b^8c^8c^8c^8} \leq \frac{a^8+a^8+b^8+b^8+b^8+c^8+c^8+c^8}{8}$ $ = \frac28 a^8 + \frac38 b^8 + \frac38 c^8$ and similarly $b^2c^3a^3 \leq \frac28 b^8 + \frac38 c^8 + \frac38 a^8$ and $c^2a^3b^3 \leq \frac28 c^8 + \frac38 a^8 + \frac38 b^8$. Summing up, we get $a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3$ $\leq \left(\frac28 a^8 + \frac38 b^8 + \frac38 c^8\right) + \left(\frac28 b^8 + \frac38 c^8 + \frac38 a^8\right)$ $ + \left(\frac28 c^8 + \frac38 a^8 + \frac38 b^8\right)$ $=a^8+b^8+c^8$, qed.. Darij

I got the same solution as Darij. Most competitions don't allow 1 line proofs by Muirhead's inequality, but the fact that the inequality holds by Muirhead guarantees a solution by AM-GM or weighted AM-GM

and if we can use amgm,we can also use sos form

Actually, this is very easy problem: WLOG let $a \geq b \geq c$. Then by Chebyshev's inequality we have $3(a^8+b^8+c^8) \geq (a^2+b^2+c^2)(a^6+b^6+c^6) \geq (ab+bc+ca)(3a^2b^2c^2)=3(ab+bc+ca)(a^2b^2c^2)$ Now after dividing everything with $3a^3b^3c^3$ we have $\frac{(a^8+b^8+c^8)}{a^3b^3c^3} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, and we're done.

Why isn't there some really troll solution yet? Ok, may I do that part... We know that $a^2+b^2+c^2 \ge ab+bc+ca$ for any reals. Thus, $a^8+b^8+c^8 \ge a^2b^4+b^4c^4+c^4a^4 \ge a^2b^4c^2+b^2c^4a^2+c^2a^4b^2 \ge a^2b^2c^2(a^2+b^2+c^2) \ge (abc)^2(ab+bc+ca)$ which is the desired result. Too easy...

Very nice solution.

$(5,-3,-3)\succ(1,1,1)$

Multiply both sides by $a^3b^3c^3$, yielding $a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3 \leq a^8+b^8+c^8$. Muirhead on $\left(8,0,0\right)\succ\left(3,3,2\right)$ and dividing by $2$ yields the desired result.

Consider the sequences $(3,3,2),(8,0,0)$.Then: $ (3,3,2)\prec (8,0,0)$ By Muirhead’s theorem we obtain: $T [3,3,2] \leq T [8,0,0]$ i.e. $2(a^2b^3c^3+a^3b^2c^3+a^3b^3c^2) \leq 2(a^8+b^8+c^8)$ $\implies a^2b^3c^3+a^3b^2c^3+a^3b^3c^2 \leq a^8+b^8+c^8$ $\implies \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{a^8 + b^8 + c^8}{a^3b^3c^3}$.$\blacksquare$

Let $a,b,c>0.$ Prove that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq\frac{a^2 + b^2+ c^2}{abc} \leq\frac{a^5 + b^5 + c^5}{a^2b^2c^2}\leq\frac{a^8 + b^8 + c^8}{a^3b^3c^3}$$