$A+B+C+D=\pi$, and $0\leq A,B,C,D \leq \frac{\pi}{2}$. Prove that $\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)\leq 2$

Question

I tried to simplify it from some ways.
(1).$\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)$
$=\left(\frac{1-\cos(2A)}{2}\right)^2+\left(\frac{1-\cos(2B)}{2}\right)^2+\left(\frac{1-\cos(2C)}{2}\right)^2+\left(\frac{1-\cos(2D)}{2}\right)^2$
$=1-\frac{1}{2}\left(\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)\right)+\frac{1}{4}\cdot\left(2+\frac{\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)}{2}\right)$
$=\frac{3}{2}-\frac{1}{2}\left(\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)\right)+\frac{1}{8}\left(\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)\right)$
(2).$\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)$
$=2\cos(A+B)\cos(A-B)+2\cos(C+D)\cos(C-D)$
since $A+B=\pi-(C+D)$, thus $\cos(A+B)=-\cos(C+D)$. It follows that
$=2\cos(A+B)\left(\cos(A-B)-\cos(C-D)\right)$
$=2\cos(A+B)\cdot\left(-2\sin(\frac{A-B+C-D}{2})\sin(\frac{A-B-C+D}{2})\right)$
$=2\cos(A+B)\cdot\left(-2\sin\left(\frac{\pi}{2}-(B+D)\right)\sin\left(\frac{\pi}{2}-(B+C)\right)\right)$
$=2\cos(A+B)\cdot\left(-2\cos(B+D)\cos(B+C)\right)$
$=-4\cos(A+B)\cos(A+C)\cos(A+D)$
(3).Similarly,$\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)$
$=2\cos(2A+2B)\cos(2A-2B)+2\cos(2C+2D)\cos(2C-2D)$
$=2\cos(2A+2B)\cos(2A+2C)\cos(2A+2D)$
Am I on the right track? Maybe it is a Jensen's-inequality problem but it seems not always concave up or down in the interval $[0,\frac{\pi}{2}]$. I am stuck here. Please help, and thank you.

Answer 1

Remark i): The desired inequality is still true without the condition $A, B, C, D \le \pi/2$.

Remark ii): According to my proof, for $A + B + C + D = \pi$, we have the following identity: \begin{align*} &2 - \sin^4 A - \sin^4 B - \sin^4 C - \sin^4 D \\ =\, & \sin^2 (A - B) \cos^2 (A + B) + \frac12(\cos (A + B) + \cos (A - B))^2\\ &\quad + 2\cos(C + D)\sin C \sin D + \sin^2 C\cos^2 C + \sin^2 D \cos^2 D. \end{align*}

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$\phantom{2}$

Problem 1: Let $A, B, C, D \ge 0$ with $A + B + C + D = \pi$. Prove that $$\sin^4 A + \sin^4 B + \sin^4 C + \sin^4 D \le 2.$$

Proof:

WLOG, assume that $A \ge B \ge C \ge D$.

Clearly, $C + D \le \pi/2$. We have (See Remark 1 at the end for details) $$\sin^2 (C + D) - \sin^2 C - \sin^2 D = 2\cos (C + D) \sin C \sin D \ge 0. \tag{1}$$ Thus, we have $$\sin^4 C + \sin^4 D \le \sin^2 C + \sin^2 D \le \sin^2 (C + D).$$

It suffices to prove that $$\sin^4 A + \sin^4 B + \sin^2(C + D) \le 2$$ or $$\sin^4 A + \sin^4 B + \sin^2(A + B) \le 2$$ which is written as (see Remark 2 at the end for details) $$\sin^2 (A - B) \cos^2 (A + B) + \frac12(\cos (A + B) + \cos (A - B))^2 \ge 0. \tag{2}$$

We are done.

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$\phantom{2}$

Remark 1: We have \begin{align*} &\sin^2(C + D) - \sin^2 C - \sin^2 D\\ =\,& \frac{1 - \cos (2C + 2D)}{2} - \frac{1 - \cos 2C}{2} - \sin^2 D\\ =\,& \frac{\cos 2C - \cos (2C + 2D)}{2} - \sin^2 D\\ =\,& \sin (2C + D) \sin D - \sin^2 D\\ =\,& (\sin (2C + D) - \sin D)\sin D\\ =\,& 2\cos (C + D) \sin C \sin D. \end{align*}

$\phantom{2}$

Remark 2: Let $u = A + B, v = A - B$ (correspondingly, $A = \frac{u + v}{2}, B = \frac{u - v}{2}$). We have \begin{align*} &2 - \sin^4 A - \sin^4 B - \sin^2(A + B)\\ =\,& 2 - \left(\frac{1 - \cos (u + v)}{2}\right)^2 - \left(\frac{1 - \cos (u - v)}{2}\right)^2 - \sin^2 u\\ =\,& \frac{3}{2} - \frac{\cos^2(u + v) + \cos^2 (u - v)}{4} + \frac{\cos(u + v) + \cos (u - v)}{2} - \sin^2 u \\ =\,& \frac{3}{2} - \frac{\cos^2 u \cos^2 v + \sin^2 u \sin^2 v}{2} + \cos u \cos v - \sin^2 u \\ =\,& - \cos^2 u \cos^2 v + \frac{1}{2}\cos^2 v + \frac{3}{2}\cos^2 u + \cos u \cos v \\ =\,& \cos^2 u \sin^2 v + \frac12(\cos u + \cos v)^2. \end{align*}

Answer 2

A bit of an ugly solution and would love a clever one.

First, let's assume wlog $D \le C \le B \le A$ and then $D+C \le \pi/2$ and we show that $\sin^4C+\sin^4D \le \sin^4(C+D)$ so we reduce the problem to three angles where we use calculus.

Noting that $\sin^2x-\sin^2y=\sin(x-y)\sin(x+y)$, one has $\sin^4(C+D)-\sin^4C=(\sin^2(C+D)+\sin^2C)\sin D \sin (2C+D)$ and since $C+D \le \pi/2$ then $D \le \pi-(2C+D)$ and $D \le 2C+D$ so $\sin D \le \sin (2C+D)=\sin (\pi-(2C+D))$ since $\sin$ is increasing on $[0, \pi/2]$. Since $\sin^2D \le \sin^2C$ by the same reason, one gets that $\sin^4D \le \sin^2C \sin D\sin (2C+D) \le (\sin^2(C+D)+\sin^2C)\sin D \sin (2C+D)$ and we are done with the reduction.

Now with $A+B+C =\pi, 0 \le A,B,C \le \pi/2$ we consider the function $f(x,y)=\sin^4x+\sin^4y+\sin^4(x+y)$ and show that it has only one critical point (at $x=y=\pi/3, (f(x,y)=27/16$) inside the triangle $0 \le x,y, \pi-x-y \le \pi/2$ while on the boundary at least one of $x,y,x+y=\pi/2$ and then if wlog $x+y=\pi/2$ we can apply the result in the first paragraph, hence $\sin^4x+\sin^4y \le \sin^4(x+y)=1$ (or directly as $\sin^4x+\sin^4y=\sin^4x+\cos^4x\le \sin^2x+\cos^2x=1$) so the maximum of $f$ on the closed triangle (which exists as $f$ continous) is $2$ so we are done!

So looking at $f_x=f_y=0$ we get $\sin^3x\cos x+\sin^3(x+y)\cos (x+y)=\sin^3y\cos y+\sin^3(x+y)\cos (x+y)=0$ or with the original notation $\sin^3A\cos A=\sin^3B\cos B=\sin^3C\cos C$

But now $g(x)=\sin^3x \cos x$ on $(0,\pi/2)$ has $g'(x)=3\sin^2x\cos^2x-\sin^4x=3\sin^2x-4\sin^4x$, so $g'$ has clearly only one zero at $x=\pi/3$ and $g'>0$ on $(0, \pi/3)$, while $g'<0$ on $(\pi/3,\pi/2)$ hence $g$ monotonic (so injective) on each of the two intervals, which implies that at least two of $A,B,C$ must be equal and wlog let $A=B=x$ hence from $f_x=f_y=0$ we get $\sin^3x\cos x=-\sin^3(2x)\cos(2x)$ which (since $2x \ge \pi/2$) gives $\cos x= -8\cos^3 x\cos (2x)$; but now we are inside $(0, \pi/2)$ so $\cos x \ne 0$ either, hence $1=-4(1+\cos 2x)\cos 2x$ or $\cos 2x=-1/2, x=\pi/3$ and our claim about critical points is proven and the problem solved!