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Question Number 39971 by math2018 last updated on 14/Jul/18
a>0,b>0, What is the minimum value of ((b^2 +2)/(a+b))+(a^2 /(ab+1)) ?
$${a}>\mathrm{0},{b}>\mathrm{0}, \\ $$$${What}\:{is}\:{the}\:{minimum}\:{value}\:{of} \\ $$$$\frac{{b}^{\mathrm{2}} +\mathrm{2}}{{a}+{b}}+\frac{{a}^{\mathrm{2}} }{{ab}+\mathrm{1}}\:\:\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18
(b^2 /(a+b))+(a^2 /(ab+1))+(2/(a+b))=x(say) using AM>GM ((a+b)/2)>(√(ab)) ((b^2 /2)/((a+b)/2))+((a^2 /2)/((ab+1)/2))+((2/2)/((a+b)/2))<((b^2 /2)/( (√(ab))))+((a^2 /2)/( (√(ab))))+((2/2)/( (√(ab))))→y(say) x<(b^2 /(2(√(ab))))+(a^2 /(2(√(ab))))+(2/(2(√(ab)))) x<((a^2 +b^2 +2)/(2(√(ab)))) x<((((a^2 +b^2 )/2)×2+2)/(2(√(ab)))) using ((a^2 +b^2 )/2)>(√(a^2 b^2 )) ((((a^2 +b^2 )/2)×2+2)/(2(√(ab))))>((2ab+2)/(2(√(ab))))→z(say) y>((ab+1)/( (√(ab)))) y>((((ab+1)/2)×2)/( (√(ab))))→z ((((ab+1)/2)×2)/( (√(ab))))>(((√(ab)) ×2)/( (√(ab)))) so z>2 given expression assumed to be x x<y y>z z>2 i can not co relate between x and z
$$\frac{{b}^{\mathrm{2}} }{{a}+{b}}+\frac{{a}^{\mathrm{2}} }{{ab}+\mathrm{1}}+\frac{\mathrm{2}}{{a}+{b}}={x}\left({say}\right) \\ $$$${using}\:{AM}>{GM} \\ $$$$\frac{{a}+{b}}{\mathrm{2}}>\sqrt{{ab}} \\ $$$$ \\ $$$$\frac{\frac{{b}^{\mathrm{2}} }{\mathrm{2}}}{\frac{{a}+{b}}{\mathrm{2}}}+\frac{\frac{{a}^{\mathrm{2}} }{\mathrm{2}}}{\frac{{ab}+\mathrm{1}}{\mathrm{2}}}+\frac{\frac{\mathrm{2}}{\mathrm{2}}}{\frac{{a}+{b}}{\mathrm{2}}}<\frac{\frac{{b}^{\mathrm{2}} }{\mathrm{2}}}{\:\sqrt{{ab}}}+\frac{\frac{{a}^{\mathrm{2}} }{\mathrm{2}}}{\:\sqrt{{ab}}}+\frac{\frac{\mathrm{2}}{\mathrm{2}}}{\:\sqrt{{ab}}}\rightarrow{y}\left({say}\right) \\ $$$${x}<\frac{{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{{ab}}}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}\sqrt{{ab}}}+\frac{\mathrm{2}}{\mathrm{2}\sqrt{{ab}}} \\ $$$${x}<\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}\sqrt{{ab}}} \\ $$$${x}<\frac{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}×\mathrm{2}+\mathrm{2}}{\mathrm{2}\sqrt{{ab}}} \\ $$$${using}\:\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}>\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\:\: \\ $$$$\frac{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}×\mathrm{2}+\mathrm{2}}{\mathrm{2}\sqrt{{ab}}}>\frac{\mathrm{2}{ab}+\mathrm{2}}{\mathrm{2}\sqrt{{ab}}}\rightarrow{z}\left({say}\right) \\ $$$${y}>\frac{{ab}+\mathrm{1}}{\:\sqrt{{ab}}} \\ $$$${y}>\frac{\frac{{ab}+\mathrm{1}}{\mathrm{2}}×\mathrm{2}}{\:\sqrt{{ab}}}\rightarrow{z} \\ $$$$ \\ $$$$\frac{\frac{{ab}+\mathrm{1}}{\mathrm{2}}×\mathrm{2}}{\:\sqrt{{ab}}}>\frac{\sqrt{{ab}}\:×\mathrm{2}}{\:\sqrt{{ab}}} \\ $$$${so}\:{z}>\mathrm{2} \\ $$$${given}\:{expression}\:{assumed}\:{to}\:{be}\:{x} \\ $$$${x}<{y} \\ $$$${y}>{z} \\ $$$${z}>\mathrm{2} \\ $$$${i}\:{can}\:{not}\:{co}\:{relate}\:{between}\:{x}\:{and}\:{z} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by math2018 last updated on 15/Jul/18
Thank you very much!
$${Thank}\:{you}\:{very}\:{much}! \\ $$
Answered by ajfour last updated on 14/Jul/18
f(a,b)=((b^2 +2)/(a+b))+(a^2 /(ab+1)) < ((b^2 +2)/(2(√(ab))))+(a^2 /(2(√(ab)))) = (((a−b)^2 +2ab+2)/(2(√(ab)))) =(((a−b)^2 )/(2(√(ab)))) +(√(ab))+(1/( (√(ab)))) the above expression has a minimum of value equal to 2 for a=b=1 hence f_(min) (a,b) ≤ 2 .
$$\:{f}\left({a},{b}\right)=\frac{{b}^{\mathrm{2}} +\mathrm{2}}{{a}+{b}}+\frac{{a}^{\mathrm{2}} }{{ab}+\mathrm{1}} \\ $$$$\:<\:\frac{{b}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}\sqrt{{ab}}}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}\sqrt{{ab}}}\:\:=\:\frac{\left({a}−{b}\right)^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}}{\mathrm{2}\sqrt{{ab}}} \\ $$$$\:\:\:=\frac{\left({a}−{b}\right)^{\mathrm{2}} }{\mathrm{2}\sqrt{{ab}}}\:+\sqrt{{ab}}+\frac{\mathrm{1}}{\:\sqrt{{ab}}} \\ $$$$\:\:{the}\:{above}\:{expression}\:{has}\:{a} \\ $$$${minimum}\:{of}\:{value}\:{equal}\:{to}\:\mathrm{2} \\ $$$${for}\:\:\:{a}={b}=\mathrm{1} \\ $$$${hence}\:\:\:{f}_{{min}} \left({a},{b}\right)\:\leqslant\:\mathrm{2}\:\:. \\ $$$$ \\ $$
Commented by math2018 last updated on 15/Jul/18
Thank you Sir!
$${Thank}\:{you}\:{Sir}! \\ $$$$ \\ $$
Answered by math2018 last updated on 15/Jul/18
1° a+b≥ab+1 ((b^2 +2)/(a+b))+(a^2 /(ab+1))≥(((a^2 +1)+(b^2 +1))/(a+b))≥((2a+2b)/(a+b))=2; 2° a+b<ab+1 ((b^2 +2)/(a+b))+(a^2 /(ab+1))>((a^2 +b^2 +2)/(ab+1))≥((2ab+2)/(ab+1))=2 hence the minimum value is 2.
$$\mathrm{1}°\:{a}+{b}\geqslant{ab}+\mathrm{1} \\ $$$$\frac{{b}^{\mathrm{2}} +\mathrm{2}}{{a}+{b}}+\frac{{a}^{\mathrm{2}} }{{ab}+\mathrm{1}}\geqslant\frac{\left({a}^{\mathrm{2}} +\mathrm{1}\right)+\left({b}^{\mathrm{2}} +\mathrm{1}\right)}{{a}+{b}}\geqslant\frac{\mathrm{2}{a}+\mathrm{2}{b}}{{a}+{b}}=\mathrm{2}; \\ $$$$\mathrm{2}°\:{a}+{b}<{ab}+\mathrm{1} \\ $$$$\frac{{b}^{\mathrm{2}} +\mathrm{2}}{{a}+{b}}+\frac{{a}^{\mathrm{2}} }{{ab}+\mathrm{1}}>\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}}{{ab}+\mathrm{1}}\geqslant\frac{\mathrm{2}{ab}+\mathrm{2}}{{ab}+\mathrm{1}}=\mathrm{2} \\ $$$${hence}\:{the}\:{minimum}\:{value}\:{is}\:\mathrm{2}. \\ $$

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