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}$$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $I know that G.M$\le$A.M and somehow i guess it must be used.But i really struggling to prove this kind of inequality.Any hints or solution will be appreciated.And i do apologize if this question is very basic.Thanks in advance.
 A: First of all, you should have $\geq $ or $\leq$ instead of $>$ and $<$ in your inequalities.  In each of them, the equality occurs if and only if $a=b=c$.
(i)  Use the Weighted AM-GM Inequality to show that
$$\frac{bc+ca+ab}{a+b+c}=\frac{b}{a+b+c}c+\frac{c}{a+b+c}a+\frac{a}{a+b+c}b\geq c^{\frac{b}{a+b+c}}a^{\frac{c}{a+b+c}}b^{\frac{a}{a+b+c}}\,.$$
Similarly,
$$\frac{bc+ca+ab}{a+b+c}=\frac{c}{a+b+c}b+\frac{a}{a+b+c}c+\frac{b}{a+b+c}a\geq b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}a^{\frac{b}{a+b+c}}\,.$$
Multiplying these two inequalities to get
$$\left(\frac{bc+ca+ab}{a+b+c}\right)^2\geq \left((bc)^a(ca)^b(ab)^c\right)^{\frac{1}{a+b+c}}\,,$$
which is equivalent to the required inequality.
(ii)  For the inequality on the right, note that
$$\frac{a^2+b^2+c^2}{a+b+c}=\frac{a}{a+b+c}a+\frac{b}{a+b+c}b+\frac{c}{a+b+c}c\geq a^{\frac{a}{a+b+c}}b^{\frac{b}{a+b+c}}c^{\frac{c}{a+b+c}}=\left(a^ab^bc^c\right)^{\frac{1}{a+b+c}}$$
by the Weighted AM-GM Inequality.  Thus,
$$\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c}\geq a^ab^bc^c$$
as desired.
For the inequality on the left, observe that
$$\frac{3}{a+b+c}=\frac{a}{a+b+c}\left(\frac{1}{a}\right)+\frac{b}{a+b+c}\left(\frac1b\right)+\frac{c}{a+b+c}\left(\frac1c\right)\,.$$
Thus, by the Weighted AM-GM Inequality,
$$\frac{3}{a+b+c} \geq \left(\frac1a\right)^{\frac{a}{a+b+c}}\left(\frac1b\right)^{\frac{b}{a+b+c}}\left(\frac{1}{c}\right)^{\frac{c}{a+b+c}}=\frac{1}{\left(a^ab^bc^c\right)^{\frac{1}{a+b+c}}}\,.$$
This is equivalent to the inequality to be proven.
A: $a)$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.
