First let , , , where are the angles of the acute triangle with sides , , . We need to show

\[\begin{align*}
3 &\ge \sum(\cos{B}+\cos{C})^2\\
&= \sum\left(\frac{(x+y)^2+(y+z)^2-(z+x)^2}{2(x+y)(y+z)}+\frac{(y+z)^2+(z+x)^2-(x+y)^2}{2(y+z)(z+x)}\right)^2\\
&= \sum\left(1+\frac{(x-z)^2-(z+x)^2}{2(x+y)(y+z)}+1+\frac{(y-x)^2-(x+y)^2}{2(y+z)(z+x)}\right)^2\\
&= 4\sum\left(1-\frac{x}{y+z}\left(\frac{z}{x+y}+\frac{y}{z+x}\right)\right)^2\\
&= 4\sum\left(1-\frac{xy(x+y)+xz(x+z)}{(x+y)(y+z)(z+x)}\right)^2\\
&= 4\sum\left(\frac{2xyz+yz(y+z)}{(x+y)(y+z)(z+x)}\right)^2
\end{align*},\]

or after expanding,
However, by the acute condition we have
which is equivalent to the desired inequality. Yay I finally managed to SOS something but I'm pretty sure what I did is equivalent to pure trig.

Edit: Haha, I forgot to do the LHS. In the spirit of overkill, it's equivalent to
which becomes, after expanding,
But

\[\begin{align*}
\sum{y^2z^2(y+z)^2}-2xyz\sum{x^3} &\ge \sum{y^2z^2(y^2+z^2)}-2xyz\sum{x^3}\\
&= \sum{x^4(y^2+z^2)}-\sum{2x^4yz}\ge0
\end{align*}\]

by AM-GM.