<s>[INST] Proving the given inequalities Q: Prove the given inequalities for positive a,b,c:$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $I know that G.M$\le$A.M and somehow i guess it must be used.But i really struggling to prove this kind of inequality.Any hints or solution will be appreciated.And i do apologize if this question is very basic.Thanks in advance.
 [/INST] $a)$We have by the weighted AM-GM inequality: $\sqrt{(ab)^{\frac{c}{a+b+c}}\cdot (bc)^{\frac{a}{a+b+c}}\cdot (ca)^{\frac{b}{a+b+c}}}\le \sqrt{\dfrac{3abc}{a+b+c}}\le \dfrac{ab+bc+ca}{a+b+c}\iff(ab+bc+ca)^2 \ge 3abc(a+b+c)\iff (ab)^2+(bc)^2+(ca)^2 \ge abc(a+b+c) $ which is true. 
$b)$ Consider $f(x) = x\ln x \implies f''(x) = \dfrac{1}{x} >0 $. Thus $f$ is convex and using $f(a)+f(b)+f(c) \ge 3f\left(\frac{a+b+c}{3}\right)$, the answer follows for the left inequality. For the right one, apply again the weighted AM-GM inequality $a^{\frac{a}{a+b+c}}\cdot b^{\frac{b}{a+b+c}}\cdot c^{\frac{c}{a+b+c}}\le \dfrac{a^2+b^2+c^2}{a+b+c}$ . QED.
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