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Let $a,b,c>0$ satisfy $\left(a+b+c\right)^2+\left(a^2+b^2+c^2\right)\le 4$. Find the minimum value of \

Nirjhor wrote: $\displaystyle\sum_{\text{cyc}} \dfrac{ab+1}{(a+b)^2}$.

$cyc$ means?

It means summing all versions of the inner expression with variables cyclically permuted. I've edited the statement for clarity.

Let $a+b=x,b+c=y,c+a=z$. The given condition becomes $4\ge x^2+y^2+z^2\quad (1)$
\[\begin{align*}S &= \sum_{cyc} \frac{(x-y+z)(y-z+x)+4}{4x^2}\\
& = \frac 3 4 + \sum_{cyc} \frac{4-(y-z)^2}{4x^2}\\
& \ge \frac 3 4 +\sum_{cyc} \frac{x^2+2yz}{4x^2}\quad [\text{Using (1)}]\\
& \ge \frac 3 4 + \frac 3 4 + \frac 2 4 \cdot 3\sqrt[3]{\frac{yz}{x^2}\cdot \frac{zx}{y^2}\cdot \frac{xy}{z^2}}\\
& =3\end{align*}\]
So the minimum is $3$ which is achieved by $a=b=c=\frac 1 {\sqrt 3}$

My solution.
\[\begin{eqnarray}
4S=\sum_{\text{cyc}}\dfrac{4ab+4}{(a+b)^2}&\ge&\sum_{\text{cyc}}\dfrac{4ab+(a+b+c)^2+(a^2+b^2+c^2)}{(a+b)^2} \\
&=& \sum_{\text{cyc}}\dfrac{2(a+b)^2+2(ab+bc+ca+c^2)}{(a+b)^2} \\
&=& 6+2\sum_{\text{cyc}}\dfrac{(b+c)(c+a)}{(a+b)^2} \\
&\ge & 6+6\sqrt[3]{\prod_{\text{cyc}}\dfrac{(b+c)(c+a)}{(a+b)^2}} =12.\\
\end{eqnarray}\] So $S\ge 3$ with equality at $a=b=c=1/\sqrt 3$.