Question

Can someone help me out !

got stuck in this question for an hour.


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Answer 1

See below

Explanation:

Considering
`$\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$`, then

`\Vert \vec{u} \cdot \vec{v}\Vert \leq \Vert\vec{u}\Vert \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$`

This is the Cauchy–Schwarz Inequality, therefore

`$\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 $`

We have the equation

`(\sin ^4 x )/(a) + (\cos^4 x )/(b) = (1)/(a+b), a,b\in\mathbb{N}`

We can use the Cauchy–Schwarz Inequality because
`a` and
`b` are greater than 0. In fact,
`a>0 \wedge b>0 \implies ab>0`. Using the Cauchy–Schwarz Inequality, we have

`(\sin ^4 x )/(a) + (\cos^4 x )/(b) =((\sin^2 x)^2)/(a)+((\cos^2 x))/(b)\geq ((\sin^2 x+\cos^2 x)^2)/(a+b) = (1)/(a+b)`

and the equation holds for

`\frac{\sin^2{x}}{a}=\frac{\cos^2{x}}{b}=(1)/(a+b)`

`\implies\quad \sin^2 x = (a)/(a+b) \text{ and }\cos^2 x = (b)/(a+b)`

Therefore, once we can write

`\sin^2 x = (a)/(a+b) \implies \sin^(4n)x = (a^(2n))/((a+b)^(2n)) \implies(\sin^(4n)x )/(a^(2n-1)) = (a^(2n))/((a+b)^(2n)\cdot a^(2n-1))`

It is the same thing for cosine, thus

`\cos^2 x = (b)/(a+b) \implies (\cos^(4n)x )/(b^(2n-1)) = (b^(2n))/((a+b)^(2n)\cdot b^(2n-1))`

Once

`(a^(2n))/((a+b)^(2n)\cdot a^(2n-1))+ (b^(2n))/((a+b)^(2n)\cdot b^(2n-1)) =(a^(2n))/((a+b)^(2n) \cdot (a^(2n))/(a) ) + (b^(2n))/((a+b)^(2n)\cdot (b^(2n))/(b) )`

`=(1)/((a+b)^(2n) \cdot (1)/(a) ) + (1)/((a+b)^(2n)\cdot (1)/(b) ) = (a)/((a+b)^(2n) ) + (b)/((a+b)^(2n) ) = (a+b)/((a+b)^(2n) )`

dividing both numerator and denominator by
`(a+b)`, we get

`(a+b)/((a+b)^(2n) ) = (1)/((a+b)^(2n-1) )`

Therefore, it is proved that

`(\sin ^(4n) x )/(a^(2n-1)) + (\cos^(4n) x )/(b^(2n-1)) = (1)/((a+b)^(2n-1)), a,b\in\mathbb{N}`