possible-or-not-a-a-b-b-gt-a-b-b-a-with-below-conditions-case-1-a-gt-b-gt-1-case-2-0-lt-b-lt-a-lt-1-

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

p o s s i b l e o r n o t ?

a^{a} + b^{b} > a^{b} + b^{a}

w i t h b e l o w c o n d i t i o n s :

c a s e 1) a > b > 1

c a s e 2) 0 < b < a < 1

Commented by prakash jain last updated on 12/Dec/17

a = 3, b = 2

27 + 4 = 31 > 9 + 8 = 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} - (a^{a} - b^{a}) < 0

\Leftrightarrow \frac{(a^{b} - b^{b}) - (a^{a} - b^{a})}{b - a} > 0 (∵ b - a < 0)

l e t f (x) = a^{x} - b^{x} (x > 0)

f^{'} (x) = a^{x} l n a - b^{x} l n b

(i) 1 < b < a

0 < l n b < l n a, 0 < b^{x} < a^{x}

∴ f^{'} (x) > 0 (\forall x > 0)

(i i) 0 < b < a < 1

l n b < l n a < 0, 0 < b^{x} < a^{x}

∴ f^{'} (x) < 0 (\forall x > 0)

M e a n V a l u e T h e o r e m

\exists c \in (b, a) s . t . f^{'} (c) = \frac{f (b) - f (a)}{b - a}

∴

1 < b < a \Rightarrow f^{'} (c) = \frac{a^{b} - b^{b} - (a^{a} - b^{a})}{b - a} > 0

0 < b < a \Rightarrow f^{'} (c) < 0

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

thank you very much dear . your

solution is so beautiful .