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Question Number 221234 by Nicholas666 last updated on 28/May/25
let 0 ≤ a,b,c, ≤ 2 , and a + b + c = 3 Prove that; (3/2) ≤ (1/(a + 1)) + (1/(b + 1)) + (1/(c +1 )) ≤ ((11)/6)
$$ \\ $$$$\:\:\mathrm{let}\:\mathrm{0}\:\leqslant\:{a},{b},{c},\:\leqslant\:\mathrm{2}\:,\:\mathrm{and}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}; \\ $$$$\:\:\frac{\mathrm{3}}{\mathrm{2}}\:\leqslant\:\frac{\mathrm{1}}{{a}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{b}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{c}\:+\mathrm{1}\:}\:\leqslant\:\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$ \\ $$
Commented by Frix last updated on 28/May/25
Let a≤b≤c Minimum at a=b=c=1 Maximum at a=0∧b=1∧c=2 (greatest possible inequality)
$$\mathrm{Let}\:{a}\leqslant{b}\leqslant{c} \\ $$$$\mathrm{Minimum}\:\mathrm{at}\:{a}={b}={c}=\mathrm{1} \\ $$$$\mathrm{Maximum}\:\mathrm{at}\:{a}=\mathrm{0}\wedge{b}=\mathrm{1}\wedge{c}=\mathrm{2} \\ $$$$\:\:\:\:\:\left(\mathrm{greatest}\:\mathrm{possible}\:\mathrm{inequality}\right) \\ $$
Answered by vnm last updated on 28/May/25
(1/(a+1))+(1/(b+1))+(1/(c+1))=3(((1/(a+1))+(1/(b+1))+(1/(c+1)))/3)≥3(((1/(a+1))∙(1/(b+1))∙(1/(c+1))))^(1/3) = (3/( (((a+1)(b+1)(c+1)))^(1/3) ))≥(3/(((a+1)+(b+1)+(c+1))/3))=(9/((a+b+c)+3))=(9/6)=(3/2) let a+1=2−t, b+1=2+t−u, c+1=2+u 0≤t,u≤1 we have to prove that (1/(2−t))+(1/(2+t−u))+(1/(2+u))≤((11)/6) or that 11(2−t)(2+t−u)(2+u)−6((2+t−u)(2+u)+(2−t)(2+u)+(2−t)(2+t−u))≥0 ...................................................... (11t−16)u^2 +(16t−11t^2 )u+16(1−t^2 )≥0 (16−11t)(−u^2 +tu+((16(1−t^2 ))/(16−11t)))≥0 16−11t>0, 0≤t≤1 u^2 −tu−((16(1−t^2 ))/(16−11t))=0 u_1 =((t−(√(t^2 +4((16(1−t^2 ))/(16−11t)))))/2)≤0 u_2 =((t+(√(t^2 +4((16(1−t^2 ))/(16−11t)))))/2) we have to prove that u_2 ≥1 ((t+(√(t^2 +4((16(1−t^2 ))/(16−11t)))))/2)≥1 t^2 +((64(1−t^2 ))/(16−11t))≥(2−t)^2 27t(1−t)≥0: TRUE u_1 ≤0, u_2 ≥1 ⇒ −u^2 +tu+((16(1−t^2 ))/(16−11t))≥0, 0≤u,t≤1
$$ \\ $$$$\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{1}}{{b}+\mathrm{1}}+\frac{\mathrm{1}}{{c}+\mathrm{1}}=\mathrm{3}\frac{\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{1}}{{b}+\mathrm{1}}+\frac{\mathrm{1}}{{c}+\mathrm{1}}}{\mathrm{3}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{a}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{{b}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{{c}+\mathrm{1}}}= \\ $$$$\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)}}\geqslant\frac{\mathrm{3}}{\frac{\left({a}+\mathrm{1}\right)+\left({b}+\mathrm{1}\right)+\left({c}+\mathrm{1}\right)}{\mathrm{3}}}=\frac{\mathrm{9}}{\left({a}+{b}+{c}\right)+\mathrm{3}}=\frac{\mathrm{9}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{let}\:{a}+\mathrm{1}=\mathrm{2}−{t},\:{b}+\mathrm{1}=\mathrm{2}+{t}−{u},\:{c}+\mathrm{1}=\mathrm{2}+{u} \\ $$$$\mathrm{0}\leqslant{t},{u}\leqslant\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}−{t}}+\frac{\mathrm{1}}{\mathrm{2}+{t}−{u}}+\frac{\mathrm{1}}{\mathrm{2}+{u}}\leqslant\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$\mathrm{or}\:\mathrm{that} \\ $$$$\mathrm{11}\left(\mathrm{2}−{t}\right)\left(\mathrm{2}+{t}−{u}\right)\left(\mathrm{2}+{u}\right)−\mathrm{6}\left(\left(\mathrm{2}+{t}−{u}\right)\left(\mathrm{2}+{u}\right)+\left(\mathrm{2}−{t}\right)\left(\mathrm{2}+{u}\right)+\left(\mathrm{2}−{t}\right)\left(\mathrm{2}+{t}−{u}\right)\right)\geqslant\mathrm{0} \\ $$$$……………………………………………… \\ $$$$\left(\mathrm{11}{t}−\mathrm{16}\right){u}^{\mathrm{2}} +\left(\mathrm{16}{t}−\mathrm{11}{t}^{\mathrm{2}} \right){u}+\mathrm{16}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{16}−\mathrm{11}{t}\right)\left(−{u}^{\mathrm{2}} +{tu}+\frac{\mathrm{16}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{16}−\mathrm{11}{t}}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{16}−\mathrm{11}{t}>\mathrm{0},\:\mathrm{0}\leqslant{t}\leqslant\mathrm{1} \\ $$$${u}^{\mathrm{2}} −{tu}−\frac{\mathrm{16}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{16}−\mathrm{11}{t}}=\mathrm{0} \\ $$$${u}_{\mathrm{1}} =\frac{{t}−\sqrt{{t}^{\mathrm{2}} +\mathrm{4}\frac{\mathrm{16}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{16}−\mathrm{11}{t}}}}{\mathrm{2}}\leqslant\mathrm{0} \\ $$$${u}_{\mathrm{2}} =\frac{{t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{4}\frac{\mathrm{16}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{16}−\mathrm{11}{t}}}}{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:{u}_{\mathrm{2}} \geqslant\mathrm{1} \\ $$$$\frac{{t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{4}\frac{\mathrm{16}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{16}−\mathrm{11}{t}}}}{\mathrm{2}}\geqslant\mathrm{1} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{64}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{16}−\mathrm{11}{t}}\geqslant\left(\mathrm{2}−{t}\right)^{\mathrm{2}} \\ $$$$\mathrm{27}{t}\left(\mathrm{1}−{t}\right)\geqslant\mathrm{0}:\:{TRUE} \\ $$$${u}_{\mathrm{1}} \leqslant\mathrm{0},\:{u}_{\mathrm{2}} \geqslant\mathrm{1}\:\Rightarrow\:−{u}^{\mathrm{2}} +{tu}+\frac{\mathrm{16}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{16}−\mathrm{11}{t}}\geqslant\mathrm{0},\:\mathrm{0}\leqslant{u},{t}\leqslant\mathrm{1} \\ $$
Answered by vnm last updated on 29/May/25
let A=a+1, B=b+1, C=c+1, 1≤A≤2 D=((B+C)/2)=((6−A)/2)=3−(A/2)≥2 ⇒ 3−D≤D−1 let B=D−t, C=D+t, 0≤t≤3−D=(A/2) (1/A)+(1/B)+(1/C)=(1/A)+(1/((3−(A/2))−t))+(1/((3−(A/2))+t))= (1/A)+((2(3−(A/2)))/((3−(A/2))^2 −t^2 ))≤(1/A)+((6−A)/((3−(A/2))^2 −((A/2))^2 ))=(1/A)+((6−A)/(3(3−A)))=(1/A)+(1/3)+(1/(3−A))= (1/3)+(3/(A(3−A))) ≤_(1≤A≤2) (1/3)+(3/2)=((11)/6)
$$ \\ $$$$\mathrm{let}\:{A}={a}+\mathrm{1},\:{B}={b}+\mathrm{1},\:{C}={c}+\mathrm{1},\:\mathrm{1}\leqslant{A}\leqslant\mathrm{2} \\ $$$${D}=\frac{{B}+{C}}{\mathrm{2}}=\frac{\mathrm{6}−{A}}{\mathrm{2}}=\mathrm{3}−\frac{{A}}{\mathrm{2}}\geqslant\mathrm{2}\:\Rightarrow\:\mathrm{3}−{D}\leqslant{D}−\mathrm{1} \\ $$$$\mathrm{let}\:{B}={D}−{t},\:{C}={D}+{t},\:\mathrm{0}\leqslant{t}\leqslant\mathrm{3}−{D}=\frac{{A}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{{B}}+\frac{\mathrm{1}}{{C}}=\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{\left(\mathrm{3}−\frac{{A}}{\mathrm{2}}\right)−{t}}+\frac{\mathrm{1}}{\left(\mathrm{3}−\frac{{A}}{\mathrm{2}}\right)+{t}}= \\ $$$$\frac{\mathrm{1}}{{A}}+\frac{\mathrm{2}\left(\mathrm{3}−\frac{{A}}{\mathrm{2}}\right)}{\left(\mathrm{3}−\frac{{A}}{\mathrm{2}}\right)^{\mathrm{2}} −{t}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{{A}}+\frac{\mathrm{6}−{A}}{\left(\mathrm{3}−\frac{{A}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{A}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{A}}+\frac{\mathrm{6}−{A}}{\mathrm{3}\left(\mathrm{3}−{A}\right)}=\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}−{A}}= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{{A}\left(\mathrm{3}−{A}\right)}\:\underset{\mathrm{1}\leqslant{A}\leqslant\mathrm{2}} {\leqslant}\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$

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