Should I use the taylor series expansion of the exponential function?
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you can Use Jenson inequality – math110 Dec 09 '15 at 15:50
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1In beginning calculus, this is a consequence of the fact that the function $e^x$ has positive second derivative, is "concave up." – André Nicolas Dec 09 '15 at 16:02
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1@math110 Jensen's inequality*. – user236182 Dec 09 '15 at 16:17
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By AM-GM,
$$a=e^x,\space b=e^y$$ $$\sqrt{ab}\le \frac{a+b}2$$ $$e^{\frac{x+y}2}\le \frac{e^x+e^y}2$$ Equality happens when $x=y$, so for $x\ne y$ only $<$ will hold.
Kay K.
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Would be even more easy be seen with $u=e^\frac x{2},;;v=e^\frac y{2}$ – Gyro Gearloose Dec 09 '15 at 15:55
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In general if $f''>0$ and $x>y$ then there exist $a\in ([x+y]/2,x)$ and $b\in (y,[x+y]/2)$ and $c\in (a.b)$ such that $$[f(x)+f(y)]/2 -f([x+y]/2)=$$ $$[f(x)-f([x+y]/2)]/2-[f([x+y]/2)-f(y)]/2=$$ $$=[(x-y)f'(a)]/4-[(x-y)f'(b)]/4=(x-y)^2f''(c)/4>0.$$
DanielWainfleet
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More generally if $f''>0$ and $r\in (0,1)$ and $x\ne y$ then $r f(x)+(1-r)f(y)>f(r x+(1-r)y)$. – DanielWainfleet Dec 09 '15 at 20:33