<s>[INST] Exploitation of Equality in Inequality. I want to ask: 
How do I exploit equalities in the solution of an inequality when equality holds for variables equal to each other?
One can take this example:

For positive real numbers that sum to $3$, prove that$$\sqrt{a + \sqrt{b^2 + c^2}} + 
\sqrt{b + \sqrt{c^2 + a^2}} + 
\sqrt{c + \sqrt{a^2 + b^2}} \ge
3\sqrt{\sqrt{2} + 1}$$

As this one which is a little tough for me, one can almost easily say that (though I am not so sure)

Equality holds for $a=b=c=1$.

But how do I exploit this in the solution? So far, I have squared both sides, but ended up only with a giant square root term, and still I can not see what can I do with this equality case.
This is not only for this one, but every time I solve an inequality and in the solution see that the equality holds for variables equal to each other, I can't still observe what happened in my solution from this equality.
Any explanation will be thankfully welcome!
 [/INST] The first step does not change something:
$$\sum_{cyc}\left(a+\sqrt{b^2+c^2}+2\sqrt{(a+\sqrt{b^2+c^2})(b+\sqrt{a^2+c^2})}\right)\geq9\sqrt2+9$$ or
$$\sum_{cyc}\left(\sqrt{b^2+c^2}+2\sqrt{(a+\sqrt{b^2+c^2})(b+\sqrt{a^2+c^2})}\right)\geq9\sqrt2+6.$$
The second step:
By C-S $$\sqrt{b^2+c^2}=\frac{1}{\sqrt2}\sqrt{(1^2+1^2)(b^2+c^2)}\geq\frac{b+c}{\sqrt2}$$ saves the case of the equality occurring: $a=b=c=1$.
It's enough to prove that:
$$\sum_{cyc}\left(\sqrt{a^2+b^2}+2\sqrt{\left(a+\frac{b+c}{\sqrt2}\right)\left(b+\frac{a+c}{\sqrt2}\right)}\right)\geq9\sqrt2+6.$$
The rest is the same: any step saves the case of the equality occurring, which gives the proof.
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