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Question Number 88752 by wiWiw last updated on 12/Apr/20

cos(𝛂)+cos(𝛃)+cos(𝛄)≤(3/2)  prove  the  inequality

$$\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\alpha}\right)+\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\beta}\right)+\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\gamma}\right)\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{prove}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{inequality}} \\ $$

Commented by TANMAY PANACEA. last updated on 12/Apr/20

Commented by TANMAY PANACEA. last updated on 12/Apr/20

another approach  point A,B and C lies on curve y=cosx  point A=(A,cosA)  point B=(B,cosB)  point C=(C,cosC)  centroid of triangle ABC is Q  Q=(((A+B+C)/3),((cosA+cosB+cosC)/3))  from figure  OP=((A+B+C)/3)  PQ=((cosA+cosB+cosC)/3)  Point R lies on y=cosx  co ordinate of point R {((A+B+C)/3),cos(((A+B+C)/3))}  from figure it is clear  PR>PQ  cos(((A+B+C)/3))>((cosA+cosB+cosC)/3)  (3/2)>cosA+cosB+cosC

$${another}\:{approach} \\ $$$${point}\:{A},{B}\:{and}\:{C}\:{lies}\:{on}\:{curve}\:{y}={cosx} \\ $$$${point}\:{A}=\left({A},{cosA}\right) \\ $$$${point}\:{B}=\left({B},{cosB}\right) \\ $$$${point}\:{C}=\left({C},{cosC}\right) \\ $$$${centroid}\:{of}\:{triangle}\:{ABC}\:{is}\:{Q} \\ $$$${Q}=\left(\frac{{A}+{B}+{C}}{\mathrm{3}},\frac{{cosA}+{cosB}+{cosC}}{\mathrm{3}}\right) \\ $$$${from}\:{figure}\:\:{OP}=\frac{{A}+{B}+{C}}{\mathrm{3}} \\ $$$${PQ}=\frac{{cosA}+{cosB}+{cosC}}{\mathrm{3}} \\ $$$${Point}\:{R}\:{lies}\:{on}\:{y}={cosx} \\ $$$${co}\:{ordinate}\:{of}\:{point}\:{R}\:\left\{\frac{{A}+{B}+{C}}{\mathrm{3}},{cos}\left(\frac{{A}+{B}+{C}}{\mathrm{3}}\right)\right\} \\ $$$${from}\:{figure}\:{it}\:{is}\:{clear} \\ $$$${PR}>{PQ} \\ $$$${cos}\left(\frac{{A}+{B}+{C}}{\mathrm{3}}\right)>\frac{{cosA}+{cosB}+{cosC}}{\mathrm{3}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}>{cosA}+{cosB}+{cosC} \\ $$$$ \\ $$$$ \\ $$

Answered by mind is power last updated on 12/Apr/20

we can assum a+b+c=π,a,b,c∈[0,(π/2)[  cos(x)′′=−cos(x)<0,  concav  ⇒(1/3)[cos(a)+cos(b)+cos(c)]≤cos(((a+b+c)/3))=cos((π/3))  ⇒cos(a)+coz(b)+cos(c)≤(3/2)

$${we}\:{can}\:{assum}\:{a}+{b}+{c}=\pi,{a},{b},{c}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right.\right. \\ $$$${cos}\left({x}\right)''=−{cos}\left({x}\right)<\mathrm{0}, \\ $$$${concav} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\left[{cos}\left({a}\right)+{cos}\left({b}\right)+{cos}\left({c}\right)\right]\leqslant{cos}\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)={cos}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow{cos}\left({a}\right)+{coz}\left({b}\right)+{cos}\left({c}\right)\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$

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