Trigonometric inequalities in a triangle What is the proof of
$~~\cos^2 A+\cos^2 B+\cos^2 C \leq 1 ~~$
in an acute triangle ?
This will be of help in finding the answer (if such exists) to finding the minimal T in any ∆ ABC when
$$T \geq \sin^k  A+ \sin^k  B+ \sin^k  C~ ,~~~~ k \geq 3$$

Let's first assume that $\cos^2A +\cos^2B+ \cos^2C=1$
Let's proceed to solve this
$\cos^2A +\cos^2B+ \cos^2C=1$
$\Rightarrow \cos^2A + \cos^2B - (1−\cos^2C)=0$
$\Rightarrow \cos^2A + cos^2B − sin^2C=0$
$\Rightarrow \cos^2A+(cos(B+C)cos(B−C))=0$
$\Rightarrow \cos^2A+(cos(π−A)cos(B−C))=0$   as $A+B+C = π$
$\Rightarrow \cos^2A−cosAcos(B−C)=0$
$\Rightarrow \cos A(cosA−cos(B−C))=0$
$\Rightarrow cosA(cos(π−(B+C))−cos(B−C))=0 $
$\Rightarrow (−cosA(B+C)−cos(B−C))=0 $
$\Rightarrow cosA(cosA(B+C)+cos(B−C))=0 $
$\Rightarrow −cosA(2cosBcosC)=0 $
$\Rightarrow 2cosAcosBcosC=0 $
$\Rightarrow cosA=0 or cosB=0 or cosC=0 $
$\Rightarrow A=\frac{π}{2} or B=\frac{π}{2} or C=\frac{π}{2} $
With this we can conclude that the triangle ABC will be a right-angled triangle if $\cos^2A +\cos^2B+ \cos^2C=1$.
Now when we say that $\cos^2A +\cos^2B+ \cos^2C \leq 1$ this means that all angles are lesser than $\frac{π}{2}$ which means that the given triangle is Acute angled.
I hope this helped you.
