Show that ${\sin^4x\over a}+{\cos^4x\over b}={1\over a+b} \implies {\sin^6x\over a^2}+{\cos^6x\over b^2}={1\over (a+b)^2}$

Question

If $${\sin^4x\over a}+{\cos^4x\over b}={1\over a+b}$$

then show that

$${\sin^6x\over a^2}+{\cos^6x\over b^2}={1\over (a+b)^2}$$

Answer 1

we only consider $ab>0$, Use Cauchy-Schwarz inequality,we have $$\left(\dfrac{\sin^4{x}}{a}+\dfrac{\cos^4{x}}{b}\right)(a+b)\ge (\sin^2{x}+\cos^2{x})^2=1$$ so $$\dfrac{\sin^4{x}}{a}+\dfrac{\cos^4{x}}{b}\ge\dfrac{1}{a+b}$$ so $$\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}=k$$ so $$\dfrac{\sin^6{x}}{a^2}+\dfrac{\cos^6{x}}{b^2}=\dfrac{1}{(a+b)^2}$$