If , α , β , γ ∈ ( 0 , 1 ) , then prove that : (√((1−^ α ).(1−^ β ). (1−^ γ ))) +(√(α^ .β^ .γ^ ))


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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
	$say α=sin^2 θ β=sin^2 φ , γ=sin^2 δ l.h.s.=cos θcos φcos δ +sin θsin φsin δ =((cos θ)/2){cos (φ+δ)+cos (φ−δ)} +((sin θ)/2){cos (φ−δ)−cos (φ+δ)} =((cos (φ+δ))/2)(cos θ−sin θ) +((cos (φ−δ))/2)(cos θ+sin θ) =((cos (φ+δ)cos (θ+(π/4)))/( (√2))) +((cos (φ−δ)sin (θ+(π/4)))/( (√2))) say cos (φ+δ)=A cos (φ−δ)=B l.h.s.=(√((A^2 /2)+(B^2 /2)))sin (θ+(π/4)+tan^(−1) (A/B)) < (√((A^2 +B^2 )/2)) < (√((1/2)+(1/2))) (=1) as A<1 , B<1$
	$s a y α = \sin^{2} θ$ $β = \sin^{2} ϕ, γ = \sin^{2} δ$ $l . h . s . = \cos θ \cos ϕ \cos δ$ $+ \sin θ \sin ϕ \sin δ$ $= \frac{\cos θ}{2} {\cos (ϕ + δ) + \cos (ϕ - δ)}$ $+ \frac{\sin θ}{2} {\cos (ϕ - δ) - \cos (ϕ + δ)}$ $= \frac{\cos (ϕ + δ)}{2} (\cos θ - \sin θ)$ $+ \frac{\cos (ϕ - δ)}{2} (\cos θ + \sin θ)$ $= \frac{\cos (ϕ + δ) \cos (θ + \frac{π}{4})}{\sqrt{2}}$ $+ \frac{\cos (ϕ - δ) \sin (θ + \frac{π}{4})}{\sqrt{2}}$ $s a y \cos (ϕ + δ) = A$ $\cos (ϕ - δ) = B$ $l . h . s . = \sqrt{\frac{A^{2}}{2} + \frac{B^{2}}{2}} \sin (θ + \frac{π}{4} + \tan^{- 1} \frac{A}{B})$ $< \sqrt{\frac{A^{2} + B^{2}}{2}} < \sqrt{\frac{1}{2} + \frac{1}{2}} (= 1)$ $a s A < 1, B < 1$
	Commented byajfour last updated on 19/Sep/22
	https://youtu.be/86aXbrp2ZG0
	Commented byajfour 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 bymnjuly1970 last updated on 19/Sep/22

	$b r a v o s i r a j f o r . . . .$ $i w i l l s e e y o u r y o u t u b e . . c e r t a i n l y$
	Commented byTawa11 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 bymnjuly1970 last updated on 19/Sep/22

	$b r a v o s i r W . . . t h a n k s a l o t$
	Commented byTawa11 last updated on 20/Sep/22

	$Great sir$
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