Prove that in any triangle $ABC$, $\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}$ I have two similar looking questions.

(1) Prove that in triangle $ABC$, $$\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}.$$
(2) If $\Delta ABC$ is an acute angled, then prove that
$$\cos^2A+\cos^2B+\cos^2C<\frac{3}{2}$$

If I apply Jensen's inequality, then $\cos^2x$ is a concave function, because its second derivative is $-2\cos 2x$ and with it being concave function
$$\cos^2A+\cos^2B+\cos^2C\leq\frac{3}{4}$$ which is not there in the question.How will we prove both of these questions. I have some intuition that in the second question, as $ABC$ is an acute angled triangle,this has something to do.
Please guide me in the right direction. Thank you.

$$y=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$
Now $\cos^2A-\sin^2B=\cos(A+B)\cos(A-B)=-\cos C\cos(A-B)$
$$\implies\cos^2C-\cos C\cos(A-B)+1-y=0$$ which is a Quadratic Equation in $\cos C$
$$\implies\cos^2(A-B)-4(1-y)\ge0\iff4y\ge4-\cos^2(A-B)=3+\sin^2(A-B)$$
$$\implies4y\ge3$$
The equality occurs if $\sin(A-B)=0\implies A=B\ \ \ \ (1)$ and $\cos C=\dfrac12\implies C=\dfrac\pi3\ \ \ \ (2)$
$(1),(2)\implies A=B=C$

$$\cos^2A+\cos^2B+\cos^2C=1+\cos^2C-\cos C\cos(A-B)$$
$$=1+\cos C[\cos C-\cos(A-B)] $$
$$=1-\cos C[\cos(A+B)+\cos(A-B)] $$
$$=1-2\cos A\cos B\cos C<1$$ if $0<A,B,C<\dfrac\pi2$
