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

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

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

$$ \\ $$$$\:\:\:{If}\:,\:\:\alpha\:,\:\beta\:,\:\gamma\:\in\:\left(\:\mathrm{0}\:\:,\:\:\mathrm{1}\:\right)\:\:,\:\:{then}\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:{prove}\:\:{that}\::\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt{\left(\mathrm{1}\overset{} {−}\alpha\:\right).\left(\mathrm{1}\overset{} {−}\beta\:\right).\:\left(\mathrm{1}\overset{} {−}\gamma\:\right)}\:+\sqrt{\overset{} {\alpha}.\overset{} {\beta}.\overset{} {\gamma}}\:\:<\:\mathrm{1} \\ $$$$\:\:\:\:\:\: \\ $$

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

$${say}\:\alpha=\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\beta=\mathrm{sin}\:^{\mathrm{2}} \phi\:,\:\gamma=\mathrm{sin}\:^{\mathrm{2}} \delta \\ $$$${l}.{h}.{s}.=\mathrm{cos}\:\theta\mathrm{cos}\:\phi\mathrm{cos}\:\delta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:\theta\mathrm{sin}\:\phi\mathrm{sin}\:\delta \\ $$$$=\frac{\mathrm{cos}\:\theta}{\mathrm{2}}\left\{\mathrm{cos}\:\left(\phi+\delta\right)+\mathrm{cos}\:\left(\phi−\delta\right)\right\} \\ $$$$\:\:+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\left\{\mathrm{cos}\:\left(\phi−\delta\right)−\mathrm{cos}\:\left(\phi+\delta\right)\right\} \\ $$$$=\frac{\mathrm{cos}\:\left(\phi+\delta\right)}{\mathrm{2}}\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:+\frac{\mathrm{cos}\:\left(\phi−\delta\right)}{\mathrm{2}}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right) \\ $$$$=\frac{\mathrm{cos}\:\left(\phi+\delta\right)\mathrm{cos}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:+\frac{\mathrm{cos}\:\left(\phi−\delta\right)\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${say}\:\:\mathrm{cos}\:\left(\phi+\delta\right)={A} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\left(\phi−\delta\right)={B} \\ $$$${l}.{h}.{s}.=\sqrt{\frac{{A}^{\mathrm{2}} }{\mathrm{2}}+\frac{{B}^{\mathrm{2}} }{\mathrm{2}}}\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}+\mathrm{tan}^{−\mathrm{1}} \frac{{A}}{{B}}\right) \\ $$$$\:\:\:<\:\sqrt{\frac{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }{\mathrm{2}}}\:<\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}\:\left(=\mathrm{1}\right) \\ $$$${as}\:\:{A}<\mathrm{1}\:\:,\:{B}<\mathrm{1} \\ $$

Commented by ajfour last updated on 19/Sep/22

https://youtu.be/86aXbrp2ZG0

Commented by ajfour last updated on 19/Sep/22

$${A}\:{small}\:{experimental}\:{educational} \\ $$$$\:{video}\:{of}\:{mine}\:{on}\:{youtube}.. \\ $$

Commented by mnjuly1970 last updated on 19/Sep/22

$${bravo}\:{sir}\:{ajfor}\:…. \\ $$$$\:\:{i}\:{will}\:{see}\:{your}\:{youtube}\:..{certainly} \\ $$

Commented by Tawa11 last updated on 20/Sep/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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

for 0<x<1: (√x)<(x)^(1/3) G.M.≤A.M. (√((1−α)(1−β)(1−γ)))+(√(αβγ)) <(((1−α)(1−β)(1−γ)))^(1/3) +((αβγ))^(1/3) ≤((1−α+1−β+1−γ)/3)+((α+β+γ)/3) =(3/3)=1

$${for}\:\mathrm{0}<{x}<\mathrm{1}:\:\sqrt{{x}}<\sqrt[{\mathrm{3}}]{{x}} \\ $$$${G}.{M}.\leqslant{A}.{M}. \\ $$$$ \\ $$$$\sqrt{\left(\mathrm{1}−\alpha\right)\left(\mathrm{1}−\beta\right)\left(\mathrm{1}−\gamma\right)}+\sqrt{\alpha\beta\gamma} \\ $$$$<\sqrt[{\mathrm{3}}]{\left(\mathrm{1}−\alpha\right)\left(\mathrm{1}−\beta\right)\left(\mathrm{1}−\gamma\right)}+\sqrt[{\mathrm{3}}]{\alpha\beta\gamma} \\ $$$$\leqslant\frac{\mathrm{1}−\alpha+\mathrm{1}−\beta+\mathrm{1}−\gamma}{\mathrm{3}}+\frac{\alpha+\beta+\gamma}{\mathrm{3}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{3}}=\mathrm{1} \\ $$

Commented by mnjuly1970 last updated on 19/Sep/22

$${bravo}\:{sir}\:{W}…{thanks}\:{alot} \\ $$

Commented by Tawa11 last updated on 20/Sep/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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