If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$ 
Question: If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$. 

My approach: Since $$\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b} \\ \implies \left(\frac{\sin^4x}{a}+\frac{\cos^4x}{b}\right)^2=\frac{1}{(a+b)^2} \\ \implies \frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}-\sin^2x\cos^2x\left(\frac{\sin^2x}{a}-\frac{\cos^2x}{b}\right)^2=\frac{1}{(a+b)^2}.$$ 
Therefore, if we can prove that $$\frac{\sin^2x}{a}-\frac{\cos^2x}{b}=0,$$ then we are done. But, I am not able to prove the same.
 A: Hint
Set $\sin^2x=s\iff\cos^2x=1-s$ in the given condition to form a quadratic equation in $s$
Solve to find $s=\sin^2x=\dfrac a{a+b}$
$\cos^2x=?$
A: We know, $~\sin^2x + \cos^2x = 1 ⇒\cos^2x = 1 - \sin^2x~$ and hence $~\cos^4x = (1 - \sin^2x)^2 = 1 + \sin^4x - 2\sin^2x~$.
Now 
$~\dfrac{\sin^4x}{a}+\dfrac{\cos^4x}{b}=\dfrac{1}{a+b}~$
$~\implies\dfrac{\sin^4x}{a} + \dfrac 1b(1 + \sin^4x - 2\sin^2x) = \dfrac{1}{a + b}~$
$~\implies \dfrac{\sin^4x}{a} + \dfrac 1b + \dfrac{\sin^4x}{b} - \dfrac{2}{b}\sin^2x = \dfrac{1}{a + b}~$
$~\implies \sin^4x\left(\dfrac 1a + \dfrac 1b\right) - \dfrac{2}{b}\sin^2x = \dfrac{1}{a + b} - \dfrac 1b~$
$~\implies\dfrac{a+b}{ab}\sin^4x- 2a\dfrac{\sin^2x}{ab} = \dfrac{b - a - b}{(a + b)b}~$
$~\implies(a + b)^2(\sin^2x)^2 - 2a(a + b) \sin^2x = -a²~$
$~\implies \{(a + b)\sin^2x\}^2 -2.a.(a + b)\sin^2x + a^2 = 0~$
$~\implies \{(a + b)\sin^2x - a\}^2 = 0~$
$~\implies \sin^2x = \dfrac{a}{a + b}~$
$~⇒1 - \sin^2x = \cos^2x = 1 - \dfrac{a}{a + b} = \dfrac{b}{a + b}~$
Therefore, $~\sin^2x = \dfrac{a}{a + b}~$ and $~\cos^2x = \dfrac{b}{a + b}~$.
Now 
$~\dfrac{\sin^6x}{a^2}+\dfrac{\cos^6x}{b^2}~$
$~=\dfrac 1{a^2}\left(\dfrac{a}{a + b}\right)^3+\dfrac 1{b^2}\left(\dfrac{b}{a + b}\right)^3~$
$~=\dfrac{a}{(a + b)^3}+\dfrac{b}{(a + b)^3}~$
$~=\dfrac{1}{(a + b)^2}~$
A: A cool question: write $\left(1+\frac{b}{a}\right)\sin(x)^4+\left(1+\frac{a}{b}\right)\cos(x)^4=1$ split the parantheses as $$(\sin(x)^2+\cos(x)^2)^2-2\cos(x)^2\sin(x)^2+\frac{a}{b}\cos(x)^4+\frac{b}{a}\sin(x)^4=1,$$ now you get  $2\cos(x)^2\sin(x)^2=\frac{a}{b}\cos(x)^4+\frac{b}{a}\sin(x)^4$ which is possible if and only if $\frac{a}{b}\cos(x)^2=\sin(x)^2$ as it is said $\cos(x)^2\left(1+\frac{a}{b}\right)=1$ you have them all in terms of $a$ and $b$. 
A: By Cauchy-- Schwarz
$$1=\left( \frac{\sin^2(x)}{\sqrt{a}}\cdot \sqrt{a}+ \frac{\cos^2(x)}{\sqrt{b}}\cdot \sqrt{b} \right)^2 \leq \left( \frac{\sin^4x}{a}+\frac{\cos^4x}{b} \right)(a+b)=1$$
Therefore, we have equality in Cauchy-Schwarz, and hence 
$$\frac{\frac{\sin^2(x)}{\sqrt{a}}}{ \sqrt{a}}=\frac{ \frac{\cos^2(x)}{\sqrt{b}}}{ \sqrt{b} }$$
which is the identity you want.
A: [This solution assumes that $x$ is real. It does not work if you allow for complex $x$.]
First, if $ a > 0  > b$ and (WLOG) $ a + b > 0$, then we have
$\frac{1}{a} < \frac{1}{a+b} = \frac{\sin^4}{a} + \frac{ \cos^4 }{b} \leq \frac{1}{a} $.
Hence, we must have (WLOG) $ a, b > 0 $.   
Now, we may apply Titu's lemma, which gives us
$\frac{ \sin^4 x}{ a} + \frac{ \cos^4 x } { b} \geq \frac{ ( \sin^2 x + \cos^2 x)^2 } { a + b }  = \frac{1}{a+b}$.  
The condition tells us that equality holds, thus we have $\frac{ \sin^2 x } { a} = \frac{ \cos^2 x } { b}$. 
