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Question Number 58964 by naka3546 last updated on 02/May/19

a + b + c = 1 a, b, c ≤ 1 Prove that (1/(a^2 + 1)) + (1/(b^2 + 1)) + (1/(c^2 + 1)) ≤ ((27)/(10))

$${a}\:+\:{b}\:+\:{c}\:\:=\:\:\mathrm{1} \\ $$$${a},\:{b},\:{c}\:\:\leqslant\:\:\mathrm{1} \\ $$$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:+\:\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:+\:\:\frac{\mathrm{1}}{{c}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:\leqslant\:\:\frac{\mathrm{27}}{\mathrm{10}} \\ $$

Answered by tanmay last updated on 02/May/19

((27)/(10))=((30−3)/(10))=(1+1+1)−((1/(10))+(1/(10))+(1/(10))) we have to prove (1/(a^2 +1))−1+(1/(b^2 +1))−1_ +(1/(c^2 +1))−1+(3/(10))<0 ((−a^2 )/(a^2 +1))+((−b^2 )/(b^2 +1))+((−c^2 )/(c^2 +1))+(3/(10)) (a−1)^2 >0 a^2 +1>2a (1/(a^2 +1))<(1/(2a)) (a^2 /(a^2 +1))<(a^2 /(2a)) −(a^2 /(a^2 +1))>−(a/2) −((a^2 /(a^2 +1))+(b^2 /(b^2 +1))+(c^2 /(c^2 +1)))>−((a/2)+(b/2)+(c/2)) −((a^2 /(a^2 +1))+(b^2 /(b^2 +1))+(c^2 /(c^2 +1)))+(3/(10))>((−1)/2)+(3/(10)) −((a^2 /(a^2 +1))+(b^2 /(b^2 +1))+(c^2 /(c^2 +1)))+(3/(10))>((−2)/(10)) right hand side is =((−1)/5)<0 so (1/(a^2 +1))+(1/(b^2 +1))+(1/(c^2 +1))≤((27)/(10)) others are requested for kind perusal please and check pls

$$\frac{\mathrm{27}}{\mathrm{10}}=\frac{\mathrm{30}−\mathrm{3}}{\mathrm{10}}=\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)−\left(\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}}\right) \\ $$$${we}\:{have}\:{to}\:{prove} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}_{} +\frac{\mathrm{1}}{{c}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}+\frac{\mathrm{3}}{\mathrm{10}}<\mathrm{0} \\ $$$$\frac{−{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{−{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{−{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{10}} \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$${a}^{\mathrm{2}} +\mathrm{1}>\mathrm{2}{a} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}<\frac{\mathrm{1}}{\mathrm{2}{a}} \\ $$$$\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}<\frac{{a}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$$$−\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}>−\frac{{a}}{\mathrm{2}} \\ $$$$−\left(\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{1}}\right)>−\left(\frac{{a}}{\mathrm{2}}+\frac{{b}}{\mathrm{2}}+\frac{{c}}{\mathrm{2}}\right) \\ $$$$−\left(\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{3}}{\mathrm{10}}>\frac{−\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{10}} \\ $$$$−\left(\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{3}}{\mathrm{10}}>\frac{−\mathrm{2}}{\mathrm{10}} \\ $$$${right}\:{hand}\:{side}\:{is}\:=\frac{−\mathrm{1}}{\mathrm{5}}<\mathrm{0} \\ $$$${so}\:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{c}^{\mathrm{2}} +\mathrm{1}}\leqslant\frac{\mathrm{27}}{\mathrm{10}} \\ $$$${others}\:{are}\:{requested}\:\:{for}\:{kind}\:{perusal}\:{please} \\ $$$${and}\:{check}\:{pls} \\ $$$$ \\ $$

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