Is there a way to prove $\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C}\leq\frac{9}{4}$ for a triangle, without Leibniz's inequality?

Question

Given a triangle with sides $a,b,c$ and internal angles $A,B,C$ I want to prove that $$\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C}\leq\frac{9}{4}$$

I can do this by using the circumradius of the triangle (proof below), but I want to know if there is an alternative that does not use the Leibniz inequality.

(The Leibniz inequality states that for a triangle with sides length $a,b,c$ and circumradius $R$, $a^{2}+b^{2}+c^{2}\leq 9R^{2}$)

By the sine rule: $$\frac{\sin{A}}{a}=\frac{\sin{B}}{b}=\frac{\sin{C}}{c}=\frac{1}{2R} \tag1$$

So $$\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C}=\frac{a^{2}+b^{2}+c^{2}}{4R^{2}} \tag2$$

And since $a^{2}+b^{2}+c^{2}\leq 9R^{2}$, it follows that

$$\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C} \leq \frac{9R^{2}}{4r^{2}}=\frac{9}{4} \tag3$$

Is there a way to do this without the Leibniz inequality? Preferably without reference to the circumradius at all...

Answer 1

Let $$\sin^2A + \sin ^2B + \sin^2C=t$$ $$\implies \sin^2A + 1 - \cos ^2 B + \sin^2 C=t$$ $$\implies1 -(\cos^2B-\sin^2A)+1-\cos^2C=t$$ $$\implies2-\cos(A+B)\cos(A-B) - \cos^2 C=t$$ As $A+B+C=π$ $$\implies2-\cos(π-C)\cos(A-B)-\cos^2C=t$$ $$\implies -\cos^2C+\cos(C)\cos(A-B) +(2-t)=0$$ This is a quadratic eqn with respect to $\cos C$. As $\cos C∈R$, $Δ≥0$. So, $$\cos^2(A-B)+4(2-t)≥0$$ $$\implies \cos^2(A-B)+8≥4t$$ As $\cos^2(A-B)_{max}=1$, $$9≥\cos^2(A-B)+8≥4t$$ $$\implies \frac{9}{4}≥t$$ $$\implies \frac{9}{4}≥\sin^2A + \sin ^2B + \sin^2C$$

Answer 2

Here is another solution using vectors: Let $E=sin^2A+sin^2B+sin^2C$

Note that $2E=3-(cos2A+cos2B+cos2C)$

lemma: for $2A + 2B +2C =2\pi, cos2A+cos2B+cos2C>= \frac {-3}{2}$

proof: consider ,3 non parallel unit vectors $\mathbf a, \mathbf b, \mathbf c$(the angles between them are 2A,2B,2C as A,B,C range from $0-to-\pi$. It is not hard to see that the expression is actually the cyclic dot product of the vectors, taken 2 at a time. ie. $\mathbf {a.b+b.c+c.a}$

but then it is a well known result that the minimum of such a product is -3/2.(actually for a proof you can consider the square of the modulus of the sum of the vectors, and write it as dot products).

Hence, we get our original expression 2E to be less than equal to $\frac 92$. Or $E<=9/4$.

Answer 3

Take $f(x,y,z)=\sin^2(x)+\sin^2(y)+\sin^2(z)$ defined over the rectangle $[0,\pi]^{3}$. Maximize this function with the constraint $x+y+z=\pi$ using Lagrange multipliers. For that, consider $g(x,y,z)=x+y+z$ and find $\lambda\in\mathbb{R}$ such that $\nabla f= \lambda \nabla g$. Then $$\nabla f=(2\sin(x)\cos(x),2\sin(y)\cos(y),2\sin(z)\cos(z))$$ and $$\nabla g=(1,1,1)$$Solve the system \begin{equation} \left\{ \begin{array}{ccc} 2\sin(x)\cos(x)=\lambda\\ 2\sin(y)\cos(y)=\lambda\\ 2\sin(z)\cos(z)=\lambda\\ x+y+z=\pi \end{array} \right. \end{equation}