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Question Number 25589 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17

possible or not ? a^a +b^b > a^b +b^a with below conditions: case 1) a>b>1 case 2) 0<b<a<1

$$\boldsymbol{{possible}}\:\boldsymbol{{or}}\:\boldsymbol{{not}}\:? \\ $$ $$\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\boldsymbol{{a}}} +\boldsymbol{{b}}^{\boldsymbol{{b}}} >\:\boldsymbol{{a}}^{\boldsymbol{{b}}} +\boldsymbol{{b}}^{\boldsymbol{{a}}} \\ $$ $$\boldsymbol{{with}}\:\boldsymbol{{below}}\:\boldsymbol{{conditions}}: \\ $$ $$\left.\:\boldsymbol{{case}}\:\mathrm{1}\right)\:\:\boldsymbol{{a}}>\boldsymbol{{b}}>\mathrm{1} \\ $$ $$\left.\boldsymbol{{case}}\:\mathrm{2}\right)\:\:\:\:\mathrm{0}<\boldsymbol{{b}}<\boldsymbol{{a}}<\mathrm{1} \\ $$

Commented byprakash jain last updated on 12/Dec/17

a=3,b=2 27+4=31>9+8=17

$${a}=\mathrm{3},{b}=\mathrm{2} \\ $$ $$\mathrm{27}+\mathrm{4}=\mathrm{31}>\mathrm{9}+\mathrm{8}=\mathrm{17} \\ $$

Commented bymoxhix last updated on 12/Dec/17

a^b +b^a <a^a +b^b , b<a ⇔a^b −b^b −(a^a −b^a )<0 ⇔(((a^b −b^b )−(a^a −b^a ))/(b−a))>0 (∵b−a<0) let f(x)=a^x −b^x (x>0) f ′(x)=a^x lna−b^x lnb (i)1<b<a 0<lnb<lna, 0<b^x <a^x ∴f ′(x)>0 (∀x>0) (ii)0<b<a<1 lnb<lna<0, 0<b^x <a^x ∴f ′(x)<0 (∀x>0) Mean Value Theorem ∃c∈(b,a)s.t.f ′(c)=((f(b)−f(a))/(b−a)) ∴ 1<b<a⇒f ′(c)=((a^b −b^b −(a^a −b^a ))/(b−a))>0 0<b<a⇒f ′(c)<0

$${a}^{{b}} +{b}^{{a}} <{a}^{{a}} +{b}^{{b}} \:\:\:,\:{b}<{a} \\ $$ $$\Leftrightarrow{a}^{{b}} −{b}^{{b}} −\left({a}^{{a}} −{b}^{{a}} \right)<\mathrm{0} \\ $$ $$\Leftrightarrow\frac{\left({a}^{{b}} −{b}^{{b}} \right)−\left({a}^{{a}} −{b}^{{a}} \right)}{{b}−{a}}>\mathrm{0}\:\:\left(\because{b}−{a}<\mathrm{0}\right) \\ $$ $$ \\ $$ $${let}\:{f}\left({x}\right)={a}^{{x}} −{b}^{{x}} \:\:\left({x}>\mathrm{0}\right) \\ $$ $${f}\:'\left({x}\right)={a}^{{x}} {lna}−{b}^{{x}} {lnb} \\ $$ $$\left({i}\right)\mathrm{1}<{b}<{a} \\ $$ $$\:\:\mathrm{0}<{lnb}<{lna},\:\mathrm{0}<{b}^{{x}} <{a}^{{x}} \\ $$ $$\:\:\therefore{f}\:'\left({x}\right)>\mathrm{0}\:\left(\forall{x}>\mathrm{0}\right) \\ $$ $$\left({ii}\right)\mathrm{0}<{b}<{a}<\mathrm{1} \\ $$ $$\:\:{lnb}<{lna}<\mathrm{0},\:\mathrm{0}<{b}^{{x}} <{a}^{{x}} \\ $$ $$\:\:\therefore{f}\:'\left({x}\right)<\mathrm{0}\:\left(\forall{x}>\mathrm{0}\right) \\ $$ $$ \\ $$ $${Mean}\:{Value}\:{Theorem} \\ $$ $$\exists{c}\in\left({b},{a}\right){s}.{t}.{f}\:'\left({c}\right)=\frac{{f}\left({b}\right)−{f}\left({a}\right)}{{b}−{a}} \\ $$ $$\therefore \\ $$ $$\:\:\mathrm{1}<{b}<{a}\Rightarrow{f}\:'\left({c}\right)=\frac{{a}^{{b}} −{b}^{{b}} −\left({a}^{{a}} −{b}^{{a}} \right)}{{b}−{a}}>\mathrm{0} \\ $$ $$\:\:\mathrm{0}<{b}<{a}\Rightarrow{f}\:'\left({c}\right)<\mathrm{0} \\ $$

Commented bybehi.8.3.4.17@gmail.com last updated on 12/Dec/17

thank you very much dear.your solution is so beautiful.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}.\mathrm{your}\: \\ $$ $$\mathrm{solution}\:\mathrm{is}\:\mathrm{so}\:\mathrm{beautiful}. \\ $$

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