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Question Number 13201 by Tinkutara last updated on 16/May/17

In a ΔABC prove that: ((sin A + sin B)/2) ≤ sin (((A + B)/2))

$$\mathrm{In}\:\mathrm{a}\:\Delta{ABC}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{sin}\:{A}\:+\:\mathrm{sin}\:\mathrm{B}}{\mathrm{2}}\:\leqslant\:\mathrm{sin}\:\left(\frac{{A}\:+\:{B}}{\mathrm{2}}\right) \\ $$

Commented by ajfour last updated on 16/May/17

kindly check if B=0 and A=−(π/3) .

$${kindly}\:{check}\:{if} \\ $$$${B}=\mathrm{0}\:{and}\:{A}=−\frac{\pi}{\mathrm{3}}\:. \\ $$

Answered by ajfour last updated on 16/May/17

A+B=π−C ((A+B)/2) = (π/2)−(C/2) ⇒ sin (((A+B)/2))=cos (C/2) > 0 sin A+sin B=2sin (((A+B)/2))cos (((A−B)/2)) ((sin A+sin B)/(2sin (((A+B)/2)))) = cos (((A−B)/2))≤1 hence ((sin A+sin B)/2) ≤ sin (((A+B)/2)) sin (((A+B)/2))>0 so inequality doesn′t change .

$${A}+{B}=\pi−{C} \\ $$$$\frac{{A}+{B}}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\:>\:\mathrm{0}\: \\ $$$$\:\mathrm{sin}\:{A}+\mathrm{sin}\:{B}=\mathrm{2sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{sin}\:{A}+\mathrm{sin}\:{B}}{\mathrm{2sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)}\:=\:\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)\leqslant\mathrm{1} \\ $$$${hence} \\ $$$$\frac{\mathrm{sin}\:{A}+\mathrm{sin}\:{B}}{\mathrm{2}}\:\leqslant\:\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)>\mathrm{0}\:\:\:{so}\:{inequality}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{doesn}'{t}\:{change}\:. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

sin(((A+B)/2))=sin(90−(C/2))=cos(C/2) sinA+sinB=2sin((A+B)/2)cos((A−B)/2)= =2sin(90−(C/2))cos(((A−B)/2))=2cos(C/2)cos((A−B)/2) ⇒((2cos(C/2)cos((A−B)/2))/2)≤cos(C/2)⇒ ⇒cos((A−B)/2)≤1 (alwyes true) ∀x∈R:cosx≤1

$${sin}\left(\frac{{A}+{B}}{\mathrm{2}}\right)={sin}\left(\mathrm{90}−\frac{{C}}{\mathrm{2}}\right)={cos}\frac{{C}}{\mathrm{2}} \\ $$$${sinA}+{sinB}=\mathrm{2}{sin}\frac{{A}+{B}}{\mathrm{2}}{cos}\frac{{A}−{B}}{\mathrm{2}}= \\ $$$$=\mathrm{2}{sin}\left(\mathrm{90}−\frac{{C}}{\mathrm{2}}\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)=\mathrm{2}{cos}\frac{{C}}{\mathrm{2}}{cos}\frac{{A}−{B}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{2}{cos}\frac{{C}}{\mathrm{2}}{cos}\frac{{A}−{B}}{\mathrm{2}}}{\mathrm{2}}\leqslant{cos}\frac{{C}}{\mathrm{2}}\Rightarrow \\ $$$$\Rightarrow{cos}\frac{{A}−{B}}{\mathrm{2}}\leqslant\mathrm{1}\:\left({alwyes}\:{true}\right)\:\:\forall{x}\in{R}:{cosx}\leqslant\mathrm{1}\: \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

sinA+sinB=(a/(2R))+(b/(2R))=((a+b)/(2R)) sin(((A+B)/2))=cos(C/2)=(√((1+cosC)/2))= =(√((1+((a^2 +b^2 −c^2 )/(2ab)))/2))=(√(((a+b+c)(a+b−c))/(4ab))) =(√((p(p−c))/(ab))) (p=((a+b+c)/2),S=((abc)/(4R))) ⇒((a+b)/(4R))≤(√((p(p−c))/(ab)))⇒(((a+b)^2 )/(16R^2 ))≤((p(p−c))/(ab))⇒ (((a+b)^2 )/((a^2 b^2 c^2 )/S^2 ))≤((p(p−c))/(ab))⇒S^2 .(((a+b)^2 )/(abc^2 ))≤p(p−c)⇒ p(p−a)(p−b)(p−c)(((a+b)^2 )/(abc^2 ))≤p(p−c)⇒ (p−a)(p−b)(a+b)^2 ≤abc^2 ⇒ (c+b−a)(c−b+a)(a+b)^2 ≤4abc^2 ⇒ (c^2 −(b−a)^2 )(a+b)^2 ≤4abc^2 ⇒ c^2 ((a+b)^2 −4ab)≤(b−a)^2 (b+a)^2 ⇒ c^2 (b−a)^2 ≤(b−a)^2 (b+a)^2 ⇒c^2 ≤(b+a)^2 ⇒ c≤b+a .■ (proved) note:if a=b⇒((sinA+sinB)/2)=sinA≤sin(((A+A)/2))=sinA.

$${sinA}+{sinB}=\frac{{a}}{\mathrm{2}{R}}+\frac{{b}}{\mathrm{2}{R}}=\frac{{a}+{b}}{\mathrm{2}{R}} \\ $$$${sin}\left(\frac{{A}+{B}}{\mathrm{2}}\right)={cos}\frac{{C}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}+{cosC}}{\mathrm{2}}}= \\ $$$$=\sqrt{\frac{\mathrm{1}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}}{\mathrm{2}}}=\sqrt{\frac{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)}{\mathrm{4}{ab}}} \\ $$$$=\sqrt{\frac{{p}\left({p}−{c}\right)}{{ab}}}\:\:\:\:\:\left({p}=\frac{{a}+{b}+{c}}{\mathrm{2}},{S}=\frac{{abc}}{\mathrm{4}{R}}\right) \\ $$$$\Rightarrow\frac{{a}+{b}}{\mathrm{4}{R}}\leqslant\sqrt{\frac{{p}\left({p}−{c}\right)}{{ab}}}\Rightarrow\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{16}{R}^{\mathrm{2}} }\leqslant\frac{{p}\left({p}−{c}\right)}{{ab}}\Rightarrow \\ $$$$\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{S}^{\mathrm{2}} }}\leqslant\frac{{p}\left({p}−{c}\right)}{{ab}}\Rightarrow{S}^{\mathrm{2}} .\frac{\left({a}+{b}\right)^{\mathrm{2}} }{{abc}^{\mathrm{2}} }\leqslant{p}\left({p}−{c}\right)\Rightarrow \\ $$$${p}\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)\frac{\left({a}+{b}\right)^{\mathrm{2}} }{{abc}^{\mathrm{2}} }\leqslant{p}\left({p}−{c}\right)\Rightarrow \\ $$$$\left({p}−{a}\right)\left({p}−{b}\right)\left({a}+{b}\right)^{\mathrm{2}} \leqslant{abc}^{\mathrm{2}} \Rightarrow \\ $$$$\left({c}+{b}−{a}\right)\left({c}−{b}+{a}\right)\left({a}+{b}\right)^{\mathrm{2}} \leqslant\mathrm{4}{abc}^{\mathrm{2}} \Rightarrow \\ $$$$\left({c}^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} \right)\left({a}+{b}\right)^{\mathrm{2}} \leqslant\mathrm{4}{abc}^{\mathrm{2}} \Rightarrow \\ $$$${c}^{\mathrm{2}} \left(\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}{ab}\right)\leqslant\left({b}−{a}\right)^{\mathrm{2}} \left({b}+{a}\right)^{\mathrm{2}} \Rightarrow \\ $$$${c}^{\mathrm{2}} \left({b}−{a}\right)^{\mathrm{2}} \leqslant\left({b}−{a}\right)^{\mathrm{2}} \left({b}+{a}\right)^{\mathrm{2}} \Rightarrow{c}^{\mathrm{2}} \leqslant\left({b}+{a}\right)^{\mathrm{2}} \Rightarrow \\ $$$${c}\leqslant{b}+{a}\:\:\:.\blacksquare\:\:\left({proved}\right) \\ $$$${note}:{if}\:\boldsymbol{{a}}=\boldsymbol{{b}}\Rightarrow\frac{{sinA}+{sinB}}{\mathrm{2}}={sinA}\leqslant{sin}\left(\frac{{A}+{A}}{\mathrm{2}}\right)={sinA}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

we can prove that: ((sinA+sinB+sinC)/3)≤sin(((A+B+C)/3))=((√3)/2) sinA+sinB+sinC≤((3(√3))/2)

$${we}\:{can}\:{prove}\:{that}: \\ $$$$\frac{{sinA}+{sinB}+{sinC}}{\mathrm{3}}\leqslant{sin}\left(\frac{{A}+{B}+{C}}{\mathrm{3}}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${sinA}+{sinB}+{sinC}\leqslant\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

(((sinA+sinB)/2))^2 ≤sin^2 (((A+B)/2))=sin^2 (90−(C/2))=cos^2 (C/2) sin^2 A+2sinAsinB+sin^2 B≤2(1+cosC)=2(1+cos(180−A−B)) 1−cos^2 A+2sinAsinB+1−cos^2 B≤2−2cosA.cosB+2sinA.sinB −cos^2 A+2cosA.cosB−cos^2 B≤0 cos^2 A−2cosA.cosB+cos^2 B≥0 (cosA−cosB)^2 ≥0 .■(proved)

$$\left(\frac{{sinA}+{sinB}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant{sin}^{\mathrm{2}} \left(\frac{{A}+{B}}{\mathrm{2}}\right)={sin}^{\mathrm{2}} \left(\mathrm{90}−\frac{{C}}{\mathrm{2}}\right)={cos}^{\mathrm{2}} \frac{{C}}{\mathrm{2}} \\ $$$${sin}^{\mathrm{2}} {A}+\mathrm{2}{sinAsinB}+{sin}^{\mathrm{2}} {B}\leqslant\mathrm{2}\left(\mathrm{1}+{cosC}\right)=\mathrm{2}\left(\mathrm{1}+{cos}\left(\mathrm{180}−{A}−{B}\right)\right) \\ $$$$\mathrm{1}−{cos}^{\mathrm{2}} {A}+\mathrm{2}{sinAsinB}+\mathrm{1}−{cos}^{\mathrm{2}} {B}\leqslant\mathrm{2}−\mathrm{2}{cosA}.{cosB}+\mathrm{2}{sinA}.{sinB} \\ $$$$−{cos}^{\mathrm{2}} {A}+\mathrm{2}{cosA}.{cosB}−{cos}^{\mathrm{2}} {B}\leqslant\mathrm{0} \\ $$$${cos}^{\mathrm{2}} {A}−\mathrm{2}{cosA}.{cosB}+{cos}^{\mathrm{2}} {B}\geqslant\mathrm{0} \\ $$$$\left({cosA}−{cosB}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\:\:.\blacksquare\left({proved}\right) \\ $$$$ \\ $$

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