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...
 A: 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.
A: I use $abc$'s method .
Let $a+b+c=3u,ab+bc+ca=3v^2,abc=w^3$ then the problem is :
$$\frac{3u3v^2}{w^3}-1\geq \left(1+\frac{9u^2-6v^2}{3v^2}\right)^{\frac{9u^2}{9u^2-6v^2}}$$
A bit of algebra and the inequality is linear in $w$ and as it's homogeneous we can assume $a=b=1$ so we need to show :
$$\Big(1+\frac{2+c^2}{1+2c}\Big)^{\frac{(2+c)^2}{2+c^2}} \leq 2\Big(1+\frac{1}{c}\Big)\Big(1+c\Big)$$
We use logarithm the inequality now is :
$$\frac{(2+c)^2}{2+c^2}\ln\Big(1+\frac{2+c^2}{1+2c}\Big)\leq \ln(2)+\ln(1+c)+\ln\Big(1+\frac{1}{c}\Big)$$
We introduce the function :
$$f\left(x\right)=\frac{(2+x)^{2}}{2+x^{2}}\ln\left(1+\frac{2+x^{2}}{1+2x}\right)-\left(\ln\left(2\right)+\ln\left(1+\frac{1}{x}\right)+\ln\left(x+1\right)\right)$$
Now I haven't an idea to conclude except that we have for $x\in(0,10]$:
$$\ln\left(2\right)+\ln\left(1+\frac{1}{x}\right)+\ln\left(x+1\right)\geq\left(\left(\ln\left(2\right)+\frac{\ln\left(2\right)}{x}+\ln\left(2\right)\ln\left(xe\right)\right)\left(\frac{x}{x+1}\right)+\frac{1}{x+1}\cdot3\ln\left(2\right)\right)\geq \frac{(2+x)^{2}}{2+x^{2}}\ln\left(1+\frac{2+x^{2}}{1+2x}\right)$$
Wich is easier I think .
Edit :
the inequality is :
$$\left(1+\frac{2\left(1+2x\right)}{2+x^{2}}\right)\ln\left(1+\frac{\left(2+x^{2}\right)}{1+2x}\right)\leq \ln\left(2\right)+\ln\left(1+\frac{1}{x}\right)+\ln\left(x+1\right)$$
Setting $a=\frac{\left(1+2x\right)}{2+x^{2}},b=\frac{1}{x},c=1+x$ :
$$\left(1+2a\right)\ln\left(1+a^{-1}\right)\leq \ln\left(1+b\right)+\ln\left(2\right)+\ln\left(c\right)$$
With the constraint $$a=\frac{\left(c+b^{-1}\right)}{2+b^{-2}}$$
Wich is easier with derivatives .
