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Question Number 175175 by Shrinava last updated on 21/Aug/22

	Answered by TheHoneyCat last updated on 23/Aug/22

	$Let me proove a slightly more general statement :$ $let x = \frac{a}{2} ⩽ y = \frac{c}{2}$ $let t \in [0, 1] : t . x + (1 - t) . y = b$ $and let us proove that$ $e^{x} + e^{- x} + e^{y} + e^{- y} - e^{t x + (1 - t) y} - e^{- t x - (1 - t) y}$ $- e^{(1 - t) x + t y} - e^{- (1 - t) x + t y} ⩾ 0$ $Notice that the problem written this way is$ $in fact equivalent to your question (exept I$ $got read of the π upper limit, it was unnecesarry)$ $Also note that the problem beeing perfectly$ $symetric in t by the transformation t (1 - t)$ $so in fact we only need to verrify for t \in [0, \frac{1}{2}]$ ${let}^{'} s go$ $if t = 0 the problem is equivalent to :$ $e^{x} + e^{- x} + e^{y} + e^{- y} - e^{y} - e^{- y} - e^{x} - e^{- x} = 0 ⩾ 0$ $so t = 0 works$ $Now {let}^{'} s show that the function of t, wich$ $we want to say positive, is increasing from$ $now on :$ $its derivative will be :$ $- (x - y) e^{t x + (1 - t) y} - (y - x) e^{- t x - (1 - t) y}$ $- (y - x) e^{(1 - t) x + t y} - (x - y) e^{- (1 - t) x - t y}$ $Considering only the first line$ $D_{1} (t) := (y - x) (e^{b} - e^{- b})$ $b ⩾ 0$ $so (e^{b} - e^{- b}) ⩾ 0$ $and y - x ⩾ 0$ $so D_{1} ⩾ 0$ $you can show the same thing for the second line$ $but everything is inverted ({y o u}^{'} l l n e e d t o u s e$ $t ⩽ 1 / 2 b u t {t h a t}^{'} s n o t a p r o b l e m a s w e s a i d$ $e a r l i e r)$ $So the function is increasing on [0, 1 / 2]$ $So it stays positive on the whole interval$ $We conclude the proof, as we said, by$ $symetry arround 1 / 2 .$ $◻$ $h o p e t h a t a n s w e r e s y o u r q u e s t i o n .$
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