Answer:
See below
Step-by-step explanation:
Considering [tex]$\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$[/tex], then
[tex]\Vert \vec{u} \cdot \vec{v}\Vert \leq \Vert\vec{u}\Vert \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$[/tex]
This is the Cauchy–Schwarz Inequality, therefore
[tex]$\left(\sum_{i=1}^{n} u_i v_i \right)^2 \leq \left(\sum_{i=1}^{n} u_i \right)^2 \left(\sum_{i=1}^{n} v_i \right)^2 $[/tex]
We have the equation
[tex]\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} = \dfrac{1}{a+b}, a,b\in\mathbb{N}[/tex]
We can use the Cauchy–Schwarz Inequality because [tex]a[/tex] and [tex]b[/tex] are greater than 0. In fact, [tex]a>0 \wedge b>0 \implies ab>0[/tex]. Using the Cauchy–Schwarz Inequality, we have
[tex]\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} =\dfrac{(\sin^2 x)^2}{a}+\dfrac{(\cos^2 x)}{b}\geq \dfrac{(\sin^2 x+\cos^2 x)^2}{a+b} = \dfrac{1}{a+b}[/tex]
and the equation holds for
[tex]\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}[/tex]
[tex]\implies\quad \sin^2 x = \dfrac{a}{a+b} \text{ and }\cos^2 x = \dfrac{b}{a+b}[/tex]
Therefore, once we can write
[tex]\sin^2 x = \dfrac{a}{a+b} \implies \sin^{4n}x = \dfrac{a^{2n}}{(a+b)^{2n}} \implies\dfrac{\sin^{4n}x }{a^{2n-1}} = \dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}[/tex]
It is the same thing for cosine, thus
[tex]\cos^2 x = \dfrac{b}{a+b} \implies \dfrac{\cos^{4n}x }{b^{2n-1}} = \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}}[/tex]
Once
[tex]\dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}+ \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}} =\dfrac{a^{2n}}{(a+b)^{2n} \cdot \dfrac{a^{2n}}{a} } + \dfrac{b^{2n}}{(a+b)^{2n}\cdot \dfrac{b^{2n}}{b} }[/tex]
[tex]=\dfrac{1}{(a+b)^{2n} \cdot \dfrac{1}{a} } + \dfrac{1}{(a+b)^{2n}\cdot \dfrac{1}{b} } = \dfrac{a}{(a+b)^{2n} } + \dfrac{b}{(a+b)^{2n} } = \dfrac{a+b}{(a+b)^{2n} }[/tex]
dividing both numerator and denominator by [tex](a+b)[/tex], we get
[tex]\dfrac{a+b}{(a+b)^{2n} } = \dfrac{1}{(a+b)^{2n-1} }[/tex]
Therefore, it is proved that
[tex]\dfrac{\sin ^{4n} x }{a^{2n-1}} + \dfrac{\cos^{4n} x }{b^{2n-1}} = \dfrac{1}{(a+b)^{2n-1}}, a,b\in\mathbb{N}[/tex]
