Prove that $(1+\frac{a^2+b^2+c^2}{ab+bc+ca})^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq (1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})$ 
Assuming $a,b,c>0$, show that
$$\Big(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Big)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq \Big(1+\frac{a}{b}\Big)\Big(1+\frac{b}{c}\Big)\Big(1+\frac{c}{a}\Big).$$

I know from CS that $ab+bc+ca \leq a^2+b^2+c^2$ and $(a+b+c)^2 \leq 3(a^2+b^2+c^2)$ so the exponent is less than 3 and the second term in parenthesis greater than $1$, but I can't manage to convert this information, might work on the right hand side but seems like I'm missing a classical inequality since I'm a very beginner in this.
I noticed this inequality is symmetrical and homogeneous, maybe assuming $a+b+c=1$ could be useful...

Let $x = \frac{a^2+b^2+c^2}{ab + bc + ca}$. Then $x \ge 1$.
Using Bernoulli inequality, we have
$$\mathrm{LHS} = (1 + x)^{1 + 2/x} = (1 + x)\Big((1 + x)^{1/x}\Big)^2
\le (1 + x)\left(1 + x\cdot \frac{1}{x}\right)^2 = 4 + 4x.$$
It suffices to prove that
$$4 + \frac{4(a^2+b^2+c^2)}{ab + bc + ca} \le \frac{(a + b)(b + c)(c + a)}{abc}$$
or (clearing the denominators)
$$(b-c)^2a^3 + (b^3 + c^3)a^2 - 2bc(b^2 + c^2)a + b^2c^2(b + c) \ge 0.$$
It suffices to prove that
$$ (b^3 + c^3)a^2 - 2bc(b^2 + c^2)a + b^2c^2(b + c) \ge 0$$
which is true since
$$(b^3 + c^3)a^2 + b^2c^2(b + c) \ge 2\sqrt{(b^3 + c^3)a^2 \cdot b^2c^2(b + c)}
\ge 2abc(b^2 + c^2)$$
where we have used AM-GM and Cauchy-Bunyakovsky-Schwarz.
We are done.
