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=(a/2) ≤ y=(c/2) let t∈[0,1] : t.x+(1−t).y=b and let us proove that e^x +e^(−x) +e^y +e^(−y) −e^(tx+(1−t)y) −e^(−tx−(1−t)y) −e^((1−t)x+ty) −e^(−(1−t)x+ty) ≥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∈[0,(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^(tx+(1−t)y) −(y−x)e^(−tx−(1−t)y) −(y−x)e^((1−t)x+ty) −(x−y)e^(−(1−t)x−ty) 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 (you′ll need to use t≤1/2 but that′s not a problem as we said earlier) 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. □ hope that answeres your question.](https://www.tinkutara.com/question/Q175204.png)
$$\mathrm{Let}\:\mathrm{me}\:\mathrm{proove}\:\mathrm{a}\:\mathrm{slightly}\:\mathrm{more}\:\mathrm{general}\:\mathrm{statement}: \\ $$$$\mathrm{let}\:{x}=\frac{{a}}{\mathrm{2}}\:\leqslant\:{y}=\frac{{c}}{\mathrm{2}}\: \\ $$$$\mathrm{let}\:{t}\in\left[\mathrm{0},\mathrm{1}\right]\::\:{t}.{x}+\left(\mathrm{1}−{t}\right).{y}={b} \\ $$$$\mathrm{and}\:\mathrm{let}\:\mathrm{us}\:\mathrm{proove}\:\mathrm{that} \\ $$$$\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} +\mathrm{e}^{{y}} +\mathrm{e}^{−{y}} −\mathrm{e}^{{tx}+\left(\mathrm{1}−{t}\right){y}} −\mathrm{e}^{−{tx}−\left(\mathrm{1}−{t}\right){y}} \\ $$$$−\mathrm{e}^{\left(\mathrm{1}−\mathrm{t}\right){x}+{ty}} −\mathrm{e}^{−\left(\mathrm{1}−{t}\right){x}+{ty}} \geqslant\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Notice}\:\mathrm{that}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{written}\:\mathrm{this}\:\mathrm{way}\:\mathrm{is} \\ $$$$\mathrm{in}\:\mathrm{fact}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{your}\:\mathrm{question}\:\left(\mathrm{exept}\:\mathrm{I}\right. \\ $$$$\left.\mathrm{got}\:\mathrm{read}\:\mathrm{of}\:\mathrm{the}\:\pi\:\mathrm{upper}\:\mathrm{limit},\:\mathrm{it}\:\mathrm{was}\:\mathrm{unnecesarry}\right) \\ $$$$ \\ $$$$\mathrm{Also}\:\mathrm{note}\:\mathrm{that}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{beeing}\:\mathrm{perfectly} \\ $$$$\mathrm{symetric}\:\mathrm{in}\:{t}\:\mathrm{by}\:\mathrm{the}\:\mathrm{transformation}\:{t} \left(\mathrm{1}−{t}\right) \\ $$$$\mathrm{so}\:\mathrm{in}\:\mathrm{fact}\:\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{verrify}\:\mathrm{for}\:{t}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$ \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{go} \\ $$$$ \\ $$$$\mathrm{if}\:{t}=\mathrm{0}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}: \\ $$$$\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} +\mathrm{e}^{{y}} +\mathrm{e}^{−{y}} −\mathrm{e}^{{y}} −\mathrm{e}^{−{y}} −\mathrm{e}^{{x}} −\mathrm{e}^{−{x}} =\mathrm{0}\geqslant\mathrm{0} \\ $$$$\mathrm{so}\:{t}=\mathrm{0}\:\mathrm{works} \\ $$$$\mathrm{Now}\:\mathrm{let}'\mathrm{s}\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{of}\:{t},\:\mathrm{wich} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{say}\:\mathrm{positive},\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{from} \\ $$$$\mathrm{now}\:\mathrm{on}: \\ $$$$\mathrm{its}\:\mathrm{derivative}\:\mathrm{will}\:\mathrm{be}: \\ $$$$−\left({x}−{y}\right)\mathrm{e}^{{tx}+\left(\mathrm{1}−{t}\right){y}} −\left({y}−{x}\right)\mathrm{e}^{−{tx}−\left(\mathrm{1}−{t}\right){y}} \\ $$$$−\left({y}−{x}\right)\mathrm{e}^{\left(\mathrm{1}−{t}\right){x}+{ty}} −\left({x}−{y}\right)\mathrm{e}^{−\left(\mathrm{1}−{t}\right){x}−{ty}} \\ $$$$ \\ $$$$\mathrm{Considering}\:\mathrm{only}\:\mathrm{the}\:\mathrm{first}\:\mathrm{line} \\ $$$$\mathrm{D}_{\mathrm{1}} \left({t}\right):=\:\left({y}−{x}\right)\left(\mathrm{e}^{{b}} −\mathrm{e}^{−{b}} \right) \\ $$$${b}\geqslant\mathrm{0} \\ $$$$\mathrm{so}\:\left(\mathrm{e}^{{b}} −\mathrm{e}^{−{b}} \right)\geqslant\mathrm{0} \\ $$$$\mathrm{and}\:{y}−{x}\geqslant\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{D}_{\mathrm{1}} \geqslant\mathrm{0} \\ $$$$ \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{show}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{line} \\ $$$$\mathrm{but}\:\mathrm{everything}\:\mathrm{is}\:\mathrm{inverted}\:\left({you}'{ll}\:{need}\:{to}\:{use}\right. \\ $$$${t}\leqslant\mathrm{1}/\mathrm{2}\:\:{but}\:{that}'{s}\:{not}\:{a}\:{problem}\:{as}\:{we}\:{said}\: \\ $$$$\left.{earlier}\right) \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{function}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{on}\:\left[\mathrm{0},\mathrm{1}/\mathrm{2}\right] \\ $$$$\mathrm{So}\:\mathrm{it}\:\mathrm{stays}\:\mathrm{positive}\:\mathrm{on}\:\mathrm{the}\:\mathrm{whole}\:\mathrm{interval} \\ $$$$\mathrm{We}\:\mathrm{conclude}\:\mathrm{the}\:\mathrm{proof},\:\mathrm{as}\:\mathrm{we}\:\mathrm{said},\:\mathrm{by}\: \\ $$$$\mathrm{symetry}\:\mathrm{arround}\:\mathrm{1}/\mathrm{2}. \\ $$$$\Box \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${hope}\:{that}\:{answeres}\:{your}\:{question}. \\ $$