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.

Solution 1:

$$y=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$

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 $0<A,B,C<\dfrac\pi2$

Solution 2:

Multiplying by $4R^2$ and exploiting the sine theorem we get:

$$ 4R^2\sum_{cyc}\cos^2 A = 12R^2-(a^2+b^2+c^2)$$ hence the first inequality is equivalent to the trivial $OH^2\geq 0$, where $O$ is the circumcenter and $H$ is the orthocenter. On the other hand, if $ABC$ is an acute-angled triangle we have that $H$ lies inside $ABC$, hence $OH^2< R^2$ and the second inequality follows.