Inequality

Let the expression be given by S.

Now, we use the results:

$a^{4}+1 \geq 2a^{2}$ by AM-GM

and

$2(b^{2}+c^{2}) \geq (b+c)^{2}$ (which follows since on simplifying this is equivalent to $(b-c)^{2} \geq 0$ which is true.)

$\implies (b^{2}+c^{2}) \geq \frac{(b+c)^{2}}{2}$

and

$\sum (\frac{2a^{2}}{b^{2}+c^{2}}) \geq 3$ which is indirectly the Nesbitt's inequality

[you can get its proof here: https://brilliant.org/wiki/nesbitts-inequality/ But a simple proof is: $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^{2}}{ab+ac}+\dfrac{b^{2}}{bc+ba}+\dfrac{c^{2}}{ca+cb}$

$\geq \dfrac{(a+b+c)^{2}}{2(ab+bc+ca)}$ (By Titu's Lemma)

$\geq \dfrac{3}{2}$ ]

[Here we have used the well-known identity $(a+b+c)^{2} \geq 3(ab+bc+ca)$ ]

Thus, we get

$2S \geq \dfrac{2a^{2}}{b^{2}+c^{2}}+\dfrac{2b^{2}}{c^{2}+a^{2}}+\dfrac{2c^{2}}{a^{2}+b^{2}} \geq 3$

And hence the minimum value is $\boxed {1.5}$ attained when $a=b=c=1$

ALITER

As a direct consequence of the AM-GM inequality,

$a^{4}+1 \geq 2a^{2}$

Further, by the Nesbitt's inequality, we know that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge \dfrac{3}{2}. \ _\square$

Thus, applying the Cauchy-Schwarz inequality we get

$(1+1+1) \left(\dfrac{a^{2}}{(b+c)^{2}}+\dfrac{b^{2}}{(c+a)^{2}}+\dfrac{c^{2}}{(a+b)^{2}}\right) \geq \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^{2} \ge \dfrac{9}{4})$

And thus we apply these 2 results and get that the minimum value of the expression is $\boxed {1.5}$

The given expression can be written as:

$[\frac{a²}{b+c}]^{2} + [\frac{b²}{a+c}]^{2} + [\frac{c²}{b+a}]^{2} + [\frac{1}{b+c}]^{2} + [\frac{1}{a+c}]^{2} + [\frac{1}{a+c}]^{2}$

Now, applying Titu's lemma , we get:

$[\frac{a²}{b+c}]^{2} + [\frac{b²}{a+c}]^{2} + [\frac{c²}{b+a}]^{2} + [\frac{1}{b+c}]^{2} + [\frac{1}{a+c}]^{2} + [\frac{1}{a+c}]^{2} \ge \frac{1}{3}[\frac{a²}{b+c} + \frac{b²}{c+a} + \frac{c²}{a+b}]^{2} + \frac{1}{3} [\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}]^{2}$

Again applying Titu's lemma inside the two brackets , we get :

$[\frac{a²}{b+c}]^{2} + [\frac{b²}{a+c}]^{2} + [\frac{c²}{b+a}]^{2} + [\frac{1}{b+c}]^{2} + [\frac{1}{a+c}]^{2} + [\frac{1}{a+c}]^{2} \ge \frac{1}{3} [\frac{1}{2}\frac{(a + b + c)^{2}}{a+b+c}]^{2} + \frac{1}{3} [\frac{1}{2}\frac{(3)^{2}}{a+b+c}]^{2}$

$[\frac{a²}{b+c}]^{2} + [\frac{b²}{a+c}]^{2} + [\frac{c²}{b+a}]^{2} + [\frac{1}{b+c}]^{2} + [\frac{1}{a+c}]^{2} + [\frac{1}{a+c}]^{2} \ge \frac{1}{12}[a + b + c]^{2} + \frac{27}{4}\frac{1}{[a+b+c]^{2}}$

Now using AM GM inequality ,we get:

$\frac{1}{12}[a + b + c]^{2} + \frac{27}{4}\frac{1}{[a+b+c]^{2}} \ge \frac{3}{2}$

Note : In all the above inequalities , equality will occur when a = b = c = 1.

Try my problem 'A Cubic Inequality'

From the form of the expression there is no reason why any of the variables $a,\; b, \; c$ should be the biggest or the smallest one. I take then $c = b = a$ . Edit: apparently it is not true and I am not able to prove it. I leave the solution just to be able to come back to this problem later. Having this, I want

$f(a) = \frac{a^4+1}{(a+a)^2} = \frac{a^4+1}{4a^2} = \frac{a^2}{4} + \frac{1}{4a^2}$

to attain the minimum value. I calculate the derivative with respect to $a$ , which is

$f'(a) = \frac{a}{2} - \frac{1}{2a^3}$ .

I want to find $a$ for which $f'(a)=0$ . I find $a=1$ . For this value the function $f(a=1)$ has an extremum. The derivative $f'(a)$ for $a<1$ is negative and for $a >1$ is positive which means that $f(a)$ has its minimum for $a = 1$ . The minimum value of the expression from the task is then

$3f(a=1) = \boxed{1.5}$ .