If-0-1-then-prove-that-1-1-1-lt-1-

Question Number 176453 by mnjuly1970 last updated on 19/Sep/22

I f, α, β, γ \in (0, 1), t h e n

p r o v e t h a t :

\sqrt{(1 \overset{}{-} α) . (1 \overset{}{-} β) . (1 \overset{}{-} γ)} + \sqrt{\overset{}{α} . \overset{}{β} . \overset{}{γ}} < 1

Answered by ajfour last updated on 19/Sep/22

s a y α = \sin^{2} θ

β = \sin^{2} ϕ, γ = \sin^{2} δ

l . h . s . = \cos θ \cos ϕ \cos δ

+ \sin θ \sin ϕ \sin δ

= \frac{\cos θ}{2} {\cos (ϕ + δ) + \cos (ϕ - δ)}

+ \frac{\sin θ}{2} {\cos (ϕ - δ) - \cos (ϕ + δ)}

= \frac{\cos (ϕ + δ)}{2} (\cos θ - \sin θ)

+ \frac{\cos (ϕ - δ)}{2} (\cos θ + \sin θ)

= \frac{\cos (ϕ + δ) \cos (θ + \frac{π}{4})}{\sqrt{2}}

+ \frac{\cos (ϕ - δ) \sin (θ + \frac{π}{4})}{\sqrt{2}}

s a y \cos (ϕ + δ) = A

\cos (ϕ - δ) = B

l . h . s . = \sqrt{\frac{A^{2}}{2} + \frac{B^{2}}{2}} \sin (θ + \frac{π}{4} + \tan^{- 1} \frac{A}{B})

< \sqrt{\frac{A^{2} + B^{2}}{2}} < \sqrt{\frac{1}{2} + \frac{1}{2}} (= 1)

a s A < 1, B < 1

Commented by ajfour last updated on 19/Sep/22

https://youtu.be/86aXbrp2ZG0

Commented by ajfour last updated on 19/Sep/22

A s m a l l e x p e r i m e n t a l e d u c a t i o n a l

v i d e o o f m i n e o n y o u t u b e . .

Commented by mnjuly1970 last updated on 19/Sep/22

b r a v o s i r a j f o r \dots .

i w i l l s e e y o u r y o u t u b e . . c e r t a i n l y

Commented by Tawa11 last updated on 20/Sep/22

Great sir

Answered by mr W last updated on 19/Sep/22

f o r 0 < x < 1 : \sqrt{x} < \sqrt[3]{x}

G . M . ⩽ A . M .

\sqrt{(1 - α) (1 - β) (1 - γ)} + \sqrt{α β γ}

< \sqrt[3]{(1 - α) (1 - β) (1 - γ)} + \sqrt[3]{α β γ}

⩽ \frac{1 - α + 1 - β + 1 - γ}{3} + \frac{α + β + γ}{3}

= \frac{3}{3} = 1

Commented by mnjuly1970 last updated on 19/Sep/22

b r a v o s i r W \dots t h a n k s a l o t

Commented by Tawa11 last updated on 20/Sep/22

Great sir

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