Question Number 25589 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17
$$\boldsymbol{{possible}}\:\boldsymbol{{or}}\:\boldsymbol{{not}}\:? \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\boldsymbol{{a}}} +\boldsymbol{{b}}^{\boldsymbol{{b}}} >\:\boldsymbol{{a}}^{\boldsymbol{{b}}} +\boldsymbol{{b}}^{\boldsymbol{{a}}} \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{below}}\:\boldsymbol{{conditions}}: \\ $$$$\left.\:\boldsymbol{{case}}\:\mathrm{1}\right)\:\:\boldsymbol{{a}}>\boldsymbol{{b}}>\mathrm{1} \\ $$$$\left.\boldsymbol{{case}}\:\mathrm{2}\right)\:\:\:\:\mathrm{0}<\boldsymbol{{b}}<\boldsymbol{{a}}<\mathrm{1} \\ $$
Commented by prakash jain last updated on 12/Dec/17
$${a}=\mathrm{3},{b}=\mathrm{2} \\ $$$$\mathrm{27}+\mathrm{4}=\mathrm{31}>\mathrm{9}+\mathrm{8}=\mathrm{17} \\ $$
Commented by moxhix last updated on 12/Dec/17
$${a}^{{b}} +{b}^{{a}} <{a}^{{a}} +{b}^{{b}} \:\:\:,\:{b}<{a} \\ $$$$\Leftrightarrow{a}^{{b}} −{b}^{{b}} −\left({a}^{{a}} −{b}^{{a}} \right)<\mathrm{0} \\ $$$$\Leftrightarrow\frac{\left({a}^{{b}} −{b}^{{b}} \right)−\left({a}^{{a}} −{b}^{{a}} \right)}{{b}−{a}}>\mathrm{0}\:\:\left(\because{b}−{a}<\mathrm{0}\right) \\ $$$$ \\ $$$${let}\:{f}\left({x}\right)={a}^{{x}} −{b}^{{x}} \:\:\left({x}>\mathrm{0}\right) \\ $$$${f}\:'\left({x}\right)={a}^{{x}} {lna}−{b}^{{x}} {lnb} \\ $$$$\left({i}\right)\mathrm{1}<{b}<{a} \\ $$$$\:\:\mathrm{0}<{lnb}<{lna},\:\mathrm{0}<{b}^{{x}} <{a}^{{x}} \\ $$$$\:\:\therefore{f}\:'\left({x}\right)>\mathrm{0}\:\left(\forall{x}>\mathrm{0}\right) \\ $$$$\left({ii}\right)\mathrm{0}<{b}<{a}<\mathrm{1} \\ $$$$\:\:{lnb}<{lna}<\mathrm{0},\:\mathrm{0}<{b}^{{x}} <{a}^{{x}} \\ $$$$\:\:\therefore{f}\:'\left({x}\right)<\mathrm{0}\:\left(\forall{x}>\mathrm{0}\right) \\ $$$$ \\ $$$${Mean}\:{Value}\:{Theorem} \\ $$$$\exists{c}\in\left({b},{a}\right){s}.{t}.{f}\:'\left({c}\right)=\frac{{f}\left({b}\right)−{f}\left({a}\right)}{{b}−{a}} \\ $$$$\therefore \\ $$$$\:\:\mathrm{1}<{b}<{a}\Rightarrow{f}\:'\left({c}\right)=\frac{{a}^{{b}} −{b}^{{b}} −\left({a}^{{a}} −{b}^{{a}} \right)}{{b}−{a}}>\mathrm{0} \\ $$$$\:\:\mathrm{0}<{b}<{a}\Rightarrow{f}\:'\left({c}\right)<\mathrm{0} \\ $$
Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}.\mathrm{your}\: \\ $$$$\mathrm{solution}\:\mathrm{is}\:\mathrm{so}\:\mathrm{beautiful}. \\ $$