Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 144951 by loveineq last updated on 30/Jun/21

Let a,b > 0 and a+b+1 = 3ab. Prove that (1) (a^2 /(a+1))+(b^2 /(b+1)) ≥ (a/(a^2 +1))+(b/(b^2 +1)) (2) (a^2 /(b+1))+(b^2 /(a+1)) ≥ (a/(b^2 +1))+(b/(a^2 +1))

$$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+\mathrm{1}\:=\:\mathrm{3}{ab}.\:\mathrm{Prove}\:\mathrm{that} \\ $$ $$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} }{{a}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} }{{b}+\mathrm{1}}\:\geqslant\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}}{{b}^{\mathrm{2}} +\mathrm{1}} \\ $$ $$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} }{{b}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} }{{a}+\mathrm{1}}\:\geqslant\:\frac{{a}}{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented byjusttry last updated on 01/Jul/21

(1) remember AM≥GM (1+a^2 )≥ 2a (1+a^2 )≥(1+a)^2 −(1+a^2 ) (1+a^2 )+(1+a^2 )≥(1+a)^2 −(1+a^2 )+(1+a^2 ) 2(a^2 +1)≥(1+a)^2 (1/((a+1)^2 ))≥(1/(2(a^2 +1))) .....(i) multply (i) with a^2 and (a+1) (a^2 /((a+1)^2 ))≥(a^2 /(2(a^2 +1)))≥(a/(2(a^2 +1))) ((a^2 (a+1))/((a+1)^2 ))≥((a(a+1))/(2(a^2 +1))) (a^2 /((a+1)))≥((a^2 +a)/(2(a^2 +1)))≥((a(√a))/((a^2 +1)))≥(a/((a^2 +1))) →AM≥GM ((a^2 +a)/2)≥a(√a) →a+b+1=3ab, choose a=1 ⇒ a(√a)≥a so (a^2 /(a+1))≥(a/(a^2 +1)) ...(ii) similary for b (b^2 /(b+1))≥(b/(b^2 +1)) ...(iii) from (ii) &(iii) (a^2 /(a+1)) + (b^2 /(b+1)) ≥ (a/(a^2 +1)) + (b/(b^2 +1))

$$\left(\mathrm{1}\right)\:{remember}\:{AM}\geqslant{GM} \\ $$ $$\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\geqslant\:\mathrm{2}{a} \\ $$ $$\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\geqslant\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −\left(\mathrm{1}+{a}^{\mathrm{2}} \right) \\ $$ $$\left(\mathrm{1}+{a}^{\mathrm{2}} \right)+\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\geqslant\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −\left(\mathrm{1}+{a}^{\mathrm{2}} \right)+\left(\mathrm{1}+{a}^{\mathrm{2}} \right) \\ $$ $$\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)\geqslant\left(\mathrm{1}+{a}\right)^{\mathrm{2}} \\ $$ $$\frac{\mathrm{1}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\geqslant\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\:.....\left({i}\right) \\ $$ $${multply}\:\left({i}\right)\:{with}\:{a}^{\mathrm{2}} {and}\:\left({a}+\mathrm{1}\right) \\ $$ $$\frac{{a}^{\mathrm{2}} }{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\geqslant\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\geqslant\frac{{a}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ $$\frac{{a}^{\mathrm{2}} \left({a}+\mathrm{1}\right)}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\geqslant\frac{{a}\left({a}+\mathrm{1}\right)}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ $$\frac{{a}^{\mathrm{2}} }{\left({a}+\mathrm{1}\right)}\geqslant\frac{{a}^{\mathrm{2}} +{a}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\geqslant\frac{{a}\sqrt{{a}}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\geqslant\frac{{a}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\:\rightarrow{AM}\geqslant{GM}\:\:\frac{{a}^{\mathrm{2}} +{a}}{\mathrm{2}}\geqslant{a}\sqrt{{a}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrow{a}+{b}+\mathrm{1}=\mathrm{3}{ab},\:{choose}\:{a}=\mathrm{1}\:\Rightarrow \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\sqrt{{a}}\geqslant{a} \\ $$ $${so} \\ $$ $$\frac{{a}^{\mathrm{2}} }{{a}+\mathrm{1}}\geqslant\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:...\left({ii}\right) \\ $$ $${similary}\:{for}\:{b}\: \\ $$ $$\frac{{b}^{\mathrm{2}} }{{b}+\mathrm{1}}\geqslant\frac{{b}}{{b}^{\mathrm{2}} +\mathrm{1}}\:...\left({iii}\right) \\ $$ $${from}\:\left({ii}\right)\:\&\left({iii}\right) \\ $$ $$\frac{{a}^{\mathrm{2}} }{{a}+\mathrm{1}}\:+\:\frac{{b}^{\mathrm{2}} }{{b}+\mathrm{1}}\:\geqslant\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{{b}}{{b}^{\mathrm{2}} +\mathrm{1}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com