if a+b+c=1.................find max value of
a^{a}b^{b}c^{c} + a^{b}b^{c}c^{a} + a^{c}b^{a}c^{b}
1Use weighted AM-GM inequality :
\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\geq (a^{a}b^{b}c^{c})^{1/(a+b+c)}=a^{a}b^{b}c^{c}
\frac{ab+bc+ca}{a+b+c}\geq (a^{b}b^{c}c^{a})^{1/(a+b+c)}=a^{b}b^{c}c^{a}
\frac{ac+ba+cb}{a+b+c}\geq (a^{c}b^{a}c^{b})^{1/(a+b+c)}=a^{c}b^{a}c^{b}
Add all these :
\frac{(a+b+c)^{2}}{a+b+c}\geq a^{a}b^{b}c^{c}+a^{b}b^{c}c^{a}+a^{c}b^{a}c^{b}
1\geq a^{a}b^{b}c^{c}+a^{b}b^{c}c^{a}+a^{c}b^{a}c^{b}
maximum value is 1
1oh god....i jus completed forgot that der is something like weithted AM-GM inequality[2][3].............thanks tht was nice[1]
62@ Bipin...
no offence... but I think I will have to now start placing you right up there with Prophet and Kaymant sir [1] You have solved a few beauties recently with such ease [1]
1:O Nishant sir, too big for a compliment
I can't even think of getting anywhere near to these infinities
and truly speaking I have learned a lot from your's and Hari sir's posts :)