<s>[INST] 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.
 [/INST] [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 $\frac{ \sin^2 x } { a} = \frac{ \cos^2 x } { b}$. 
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